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Why Hurt Yourself?

By snob321 Apr 17, 2013 19879 Words
POTENTIAL ENERGY AND ENERGY CONSERVATION

7

7.1.

IDENTIFY: U grav = mgy so ΔU grav = mg ( y2 − y1 ) SET UP: + y is upward. EXECUTE: (a) ΔU = (75 kg)(9.80 m/s 2 )(2400 m − 1500 m) = +6.6 × 105 J (b) ΔU = (75 kg)(9.80 m/s2 )(1350 m − 2400 m) = −7.7 × 105 J EVALUATE: U grav increases when the altitude of the object increases.

7.2.

IDENTIFY: The change in height of a jumper causes a change in their potential energy. SET UP: Use ΔU grav = mg ( yf − yi ). EXECUTE: ΔU grav = (72 kg)(9.80 m/s2 )(0.60 m) = 420 J. EVALUATE: This gravitational potential energy comes from elastic potential energy stored in the jumper’s tensed muscles. IDENTIFY: Use the free-body diagram for the bag and Newton's first law to find the force the worker applies. Since the bag starts and ends at rest, K 2 − K1 = 0 and Wtot = 0. SET UP: A sketch showing the initial and final positions of the bag is given in Figure 7.3a. sin φ = 2.0 m 3. 5 m and φ = 34.85°. The free-body diagram is given in Figure 7.3b. F is the horizontal force applied by the worker. In the calculation of U grav take + y upward and y = 0 at the initial position of the bag.

7.3.

EXECUTE: (a) ΣFy = 0 gives T cos φ = mg and ΣFx = 0 gives F = T sin φ . Combining these equations to

eliminate T gives F = mg tan φ = (120 kg)(9.80 m/s 2 ) tan 34.85° = 820 N. (b) (i) The tension in the rope is radial and the displacement is tangential so there is no component of T in the direction of the displacement during the motion and the tension in the rope does no work. (ii) Wtot = 0 so Wworker = −Wgrav = U grav,2 − U grav,1 = mg ( y2 − y1 ) = (120 kg)(9.80 m/s 2 )(0.6277 m) = 740 J. EVALUATE: The force applied by the worker varies during the motion of the bag and it would be difficult to calculate Wworker directly.

Figure 7.3
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7-1

7-2 7.4.

Chapter 7 IDENTIFY: The energy from the food goes into the increased gravitational potential energy of the hiker. We must convert food calories to joules. SET UP: The change in gravitational potential energy is ΔU grav = mg ( yf − yi ), while the increase in

kinetic energy is negligible. Set the food energy, expressed in joules, equal to the mechanical energy developed. EXECUTE: (a) The food energy equals mg ( yf − yi ), so = 920 m. (65 kg)(9.80 m/s 2 ) (b) The mechanical energy would be 20% of the results of part (a), so Δy = (0.20)(920 m) = 180 m. EVALUATE: Since only 20% of the food calories go into mechanical energy, the hiker needs much less of climb to turn off the calories in the bar. IDENTIFY and SET UP: Use energy methods. Points 1 and 2 are shown in Figure 7.5. 2 (a) K1 + U1 + Wother = K 2 + U 2 . Solve for K 2 and then use K 2 = 1 mv2 to obtain v2 . 2

yf − yi =

(140 food calories)(4186 J/1 food calorie)

7.5.

Wother = 0 (The only force on the ball while
it is in the air is gravity.) 2 2 K1 = 1 mv1 ; K 2 = 1 mv2 2 2

U1 = mgy1, y1 = 22.0 m U 2 = mgy2 = 0, since y2 = 0 for our choice of coordinates. Figure 7.5 EXECUTE:
1 mv 2 1 2 2 + mgy1 = 1 mv2 2
2 v2 = v1 + 2 gy1 = (12.0 m/s) 2 + 2(9.80 m/s 2 )(22.0 m) = 24.0 m/s

EVALUATE: The projection angle of 53.1° doesn’t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity. (b) Nothing changes in the calculation. The expression derived in part (a) for v2 is independent of the

angle, so v2 = 24.0 m/s, the same as in part (a).
7.6. (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect. IDENTIFY: The normal force does no work, so only gravity does work and Eq. (7.4) applies. SET UP: K1 = 0. The crate’s initial point is at a vertical height of d sin α above the bottom of the ramp. 2 EXECUTE: (a) y2 = 0, y1 = d sin α . K1 + U grav,1 = K 2 + U grav,2 gives U grav,1 = K 2 . mgd sin α = 1 mv2 2

and v2 = 2 gd sin α .
2 (b) y1 = 0, y2 = −d sin α . K1 + U grav,1 = K 2 + U grav,2 gives 0 = K 2 + U grav,2 . 0 = 1 mv2 + (−mgd sin α ) 2

and v2 = 2 gd sin α , the same as in part (a).
(c) The normal force is perpendicular to the displacement and does no work. EVALUATE: When we use U grav = mgy we can take any point as y = 0 but we must take + y to be 7.7.

upward. IDENTIFY: The take-off kinetic energy of the flea goes into gravitational potential energy. SET UP: Use K f + U f = Ki + U i . Let yi = 0 and yf = h and note that U i = 0 while K f = 0 at the maximum height. Consequently, conservation of energy becomes mgh = 1 mvi 2 . 2 EXECUTE: (a) vi = 2 gh = 2(9.80 m/s 2 )(0.20 m) = 2.0 m/s.

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Potential Energy and Energy Conservation (b) Ki = mgh = (0.50 × 10−6 kg)(9.80 m/s 2 )(0.20 m) = 9.8 × 10−7 J. The kinetic energy per kilogram is

7-3

Ki 9.8 × 10−7 J = = 2.0 J/kg. m 0.50 × 10−6 kg
⎛ 2.0 m ⎞ ⎛l ⎞ (c) The human can jump to a height of hh = hf ⎜ h ⎟ = (0.20 m) ⎜ ⎟ = 200 m. To attain this −3 ⎜ ⎟ ⎝ lf ⎠ ⎝ 2.0 × 10 m ⎠

height, he would require a takeoff speed of: vi = 2 gh = 2(9.80 m/s2 )(200 m) = 63 m/s. Ki = gh = (9.80 m/s 2 )(0.60 m) = 5.9 J/kg. m (e) EVALUATE: The flea stores the energy in its tensed legs. IDENTIFY and SET UP: Apply Eq. (7.7) and consider how each term depends on the mass. EXECUTE: The speed is v and the kinetic energy is 4K. The work done by friction is proportional to the normal force, and hence to the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four times the mass, the kinetic energy is quadrupled. EVALUATE: The same result is obtained if we apply ΣF = ma to the motion. Each force is proportional to m and m divides out, so a is independent of m. IDENTIFY: Wtot = K B − K A . The forces on the rock are gravity, the normal force and friction.

(d) The human’s kinetic energy per kilogram is

7.8.

7.9.

SET UP: Let y = 0 at point B and let + y be upward. y A = R = 0.50 m. The work done by friction is

negative; W f = −0.22 J. K A = 0. The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of the rock at this point is arad = v 2 /R, upward. EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work. (ii) Wgrav = U grav ,A − U grav ,B = mgy A = (0.20 kg)(9.80 m/s 2 )(0.50 m) = 0.98 J. (b) Wtot = Wn + W f + Wgrav = 0 + (−0.22 J) + 0.98 J = 0.76 J. Wtot = K B − K A gives 1 mv 2 B 2

= Wtot .

vB =

2Wtot 2(0.76 J) = = 2.8 m/s. m 0.20 kg

(c) Gravity is constant and equal to mg. n is not constant; it is zero at A and not zero at B. Therefore, f k = μ k n is also not constant. (d) ΣFy = ma y applied to Figure 7.9 gives n − mg = marad .

⎛ ⎛ v2 ⎞ [2.8 m/s]2 ⎞ n = m ⎜ g + ⎟ = (0.20 kg) ⎜ 9.80 m/s 2 + ⎟ = 5.1 N. ⎜ ⎟ ⎜ R⎠ 0.50 m ⎟ ⎝ ⎝ ⎠ EVALUATE: In the absence of friction, the speed of the rock at point B would be 2 gR = 3.1 m/s. As the rock slides through point B, the normal force is greater than the weight mg = 2.0 N of the rock.

Figure 7.9

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7-4 7.10.

Chapter 7 IDENTIFY: The potential energy is transformed into kinetic energy which is then imparted to the bone. SET UP: The initial gravitational potential energy must be absorbed by the leg bones. U i = mgh. = 0.68 m = 68 cm. (60 kg)(9.80 m/s 2 ) (b) EVALUATE: They flex when they land and their joints and muscles absorb most of the energy. (c) EVALUATE: Their bones are more fragile so can absorb less energy without breaking and their muscles and joints are weaker and less flexible and therefore less able to absorb energy. IDENTIFY: Apply Eq. (7.7) to the motion of the car. SET UP: Take y = 0 at point A. Let point 1 be A and point 2 be B.

EXECUTE: (a) mgh = 2(200 J), so h =

400 J

7.11.

K1 + U1 + Wother = K 2 + U 2
EXECUTE: U1 = 0, U 2 = mg (2 R ) = 28,224 J, Wother = W f
2 2 K1 = 1 mv1 = 37,500 J, K 2 = 1 mv2 = 3840 J 2 2

The work-energy relation then gives W f = K 2 + U 2 − K1 = −5400 J. EVALUATE: Friction does negative work. The final mechanical energy ( K 2 + U 2 = 32 ,064 J) is less than

the initial mechanical energy ( K1 + U1 = 37,500 J) because of the energy removed by friction work. 7.12. IDENTIFY: Only gravity does work, so apply Eq. (7.5). 2 SET UP: v1 = 0, so 1 mv2 = mg ( y1 − y2 ). 2 EXECUTE: Tarzan is lower than his original height by a distance y1 − y2 = l (cos30° − cos 45°) so his

speed is v = 2 gl (cos30° − cos 45°) = 7.9 m/s, a bit quick for conversation. EVALUATE: The result is independent of Tarzan’s mass. 7.13.

y1 = 0 y2 = (8.00 m)sin 36.9° y2 = 4.80 m

Figure 7.13a (a) IDENTIFY and SET UP: F is constant so Eq. (6.2) can be used. The situation is sketched in Figure 7.13a. EXECUTE: WF = ( F cos φ ) s = (110 N)(cos0°)(8.00 m) = 880 J EVALUATE: F is in the direction of the displacement and does positive work. (b) IDENTIFY and SET UP: Calculate W using Eq. (6.2) but first must calculate the friction force. Use the freebody diagram for the oven sketched in Figure 7.13b to calculate the normal force n; then the friction force can be calculated from f k = μ k n. For this calculation use coordinates parallel and perpendicular to the incline. EXECUTE: ΣFy = ma y

n − mg cos36.9° = 0 n = mg cos36.9° f k = μ k n = μ k mg cos36.9° f k = (0.25)(10.0 kg)(9.80 m/s 2 )cos36.9° = 19.6 N

Figure 7.13b
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Potential Energy and Energy Conservation

7-5

W f = ( f k cos φ ) s = (19.6 N)(cos180°)(8.00 m) = −157 J EVALUATE: Friction does negative work. (c) IDENTIFY and SET UP: U = mgy; take y = 0 at the bottom of the ramp. EXECUTE: ΔU = U 2 − U1 = mg ( y2 − y1 ) = (10.0 kg)(9.80 m/s 2 )(4.80 m − 0) = 470 J EVALUATE: The object moves upward and U increases. (d) IDENTIFY and SET UP: Use Eq. (7.7). Solve for ΔK . EXECUTE: K1 + U1 + Wother = K 2 + U 2

ΔK = K 2 − K1 = U1 − U 2 + Wother ΔK = Wother − ΔU

Wother = WF + W f = 880 J − 157 J = 723 J
ΔU = 470 J Thus ΔK = 723 J − 470 J = 253 J. EVALUATE: Wother is positive. Some of Wother goes to increasing U and the rest goes to increasing K. (e) IDENTIFY: Apply ΣF = ma to the oven. Solve for a and then use a constant acceleration equation to calculate v2 . SET UP: We can use the free-body diagram that is in part (b): ΣFx = ma x

F − f k − mg sin 36.9° = ma
EXECUTE: a =

F − f k − mg sin 36.9° 110 N − 19.6 N − (10 kg)(9.80 m/s 2 )sin 36.9° = = 3.16 m/s 2 m 10.0 kg

SET UP: v1x = 0, a x = 3.16 m/s 2 , x − x0 = 8.00 m, v2 x = ? 2 2 v2 x = v1x + 2a x ( x − x0 )

EXECUTE: v2 x = 2a x ( x − x0 ) = 2(3.16 m/s 2 )(8.00 m) = 7.11 m/s 2 Then ΔK = K 2 − K1 = 1 mv2 = 1 (10.0 kg)(7.11 m/s)2 = 253 J. 2 2

7.14.

EVALUATE: This agrees with the result calculated in part (d) using energy methods. IDENTIFY: Use the information given in the problem with F = kx to find k. Then U el = 1 kx 2 . 2 SET UP: x is the amount the spring is stretched. When the weight is hung from the spring, F = mg . EXECUTE: k =

F mg (3.15 kg)(9.80 m/s 2 ) = = = 2205 N/m. x x 0.1340 m − 0.1200 m

2U el 2(10.0 J) =± = ±0.0952 m = ±9.52 cm. The spring could be either stretched 9.52 cm or k 2205 N/m compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm + 9.52 cm = 21.52 cm. If it were compressed, the total length of the spring would be 12.00 cm − 9.52 cm = 2.48 cm. EVALUATE: To stretch or compress the spring 9.52 cm requires a force F = kx = 210 N. x=± 7.15. IDENTIFY: Apply U el = 1 kx 2 . 2 SET UP: kx = F , so U = 1 Fx, where F is the magnitude of force required to stretch or compress the 2

spring a distance x. EXECUTE: (a) (1/2)(800 N)(0.200 m) = 80.0 J. (b) The potential energy is proportional to the square of the compression or extension; (80.0 J) (0.050 m/0.200 m)2 = 5.0 J. EVALUATE: We could have calculated k = F 800 N = = 4000 N/m and then used U el = 1 kx 2 directly. 2 x 0.200 m

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7-6 7.16.

Chapter 7 IDENTIFY: We treat the tendon like a spring and apply Hooke’s law to it. Knowing the force stretching the tendon and how much it stretched, we can find its force constant. SET UP: Use Fon tendon = kx. In part (a), Fon tendon equals mg, the weight of the object suspended from it.

In part(b), also apply U el = 1 kx 2 to calculate the stored energy. 2

7.17.

Fon tendon (0.250 kg)(9.80 m/s 2 ) = = 199 N/m. x 0.0123 m F 138 N (b) x = on tendon = = 0.693m = 69.3 cm; U el = 1 (199 N/m)(0.693 m) 2 = 47.8 J. 2 k 199 N/m EVALUATE: The 250 g object has a weight of 2.45 N. The 138 N force is much larger than this and stretches the tendon a much greater distance. IDENTIFY: Apply U el = 1 kx 2 . 2 EXECUTE: (a) k =

2 SET UP: U 0 = 1 kx0 . x is the distance the spring is stretched or compressed. 2

2 EXECUTE: (a) (i) x = 2 x0 gives U el = 1 k (2 x0 ) 2 = 4( 1 kx0 ) = 4U 0 . (ii) x = x0 /2 gives 2 2 2 U el = 1 k ( x0 /2) 2 = 1 ( 1 kx0 ) = U 0 /4. 2 4 2

(b) (i) U = 2U 0 gives x = x0 / 2.
7.18.

1 kx 2 2

2 = 2( 1 kx0 ) and x = x0 2. (ii) U = U 0 /2 gives 2

1 kx 2 2

2 = 1 ( 1 kx0 ) and 2 2

EVALUATE: U is proportional to x 2 and x is proportional to U . IDENTIFY: Apply Eq. (7.13). SET UP: Initially and at the highest point, v = 0, so K1 = K 2 = 0. Wother = 0. EXECUTE: (a) In going from rest in the slingshot’s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; U = mgy = (10 × 10−3 kg)(9.80 m/s 2 ) (22.0 m) = 2.16 J. (b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m. (c) The lack of air resistance and no deformation of the rubber band are two possible assumptions. EVALUATE: The potential energy stored in the rubber band depends on k for the rubber band and the maximum distance it is stretched. IDENTIFY and SET UP: Use energy methods. There are changes in both elastic and gravitational potential energy; elastic; U = 1 kx 2 , gravitational: U = mgy. 2 2U 2(3.20 J) = = 0.0632 m = 6.32 cm 1600 N/m k (b) Points 1 and 2 in the motion are sketched in Figure 7.19.

7.19.

EXECUTE: (a) U = 1 kx 2 so x = 2

K1 + U1 + Wother = K 2 + U 2 Wother = 0 (Only work is that done by gravity and spring force) K1 = 0, K 2 = 0 y = 0 at final position of book U1 = mg (h + d ), U 2 = 1 kd 2 2 Figure 7.19

0 + mg (h + d ) + 0 = 1 kd 2 2
The original gravitational potential energy of the system is converted into potential energy of the compressed spring.

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Potential Energy and Energy Conservation
1 kd 2 2

7-7

− mgd − mgh = 0

⎞ 1⎛ ⎛1 ⎞ d = ⎜ mg ± (mg ) 2 + 4 ⎜ k ⎟ ( mgh) ⎟ ⎜ ⎟ 2 ⎠ k⎝ ⎝ ⎠ 1 d must be positive, so d = mg + (mg ) 2 + 2kmgh k 1 d= (1.20 kg)(9.80 m/s 2 ) + 1600 N/m

(

)

7.20.

((1.20 kg)(9.80 m/s 2 ))2 + 2(1600 N/m)(1.20 kg)(9.80 m/s 2 )(0.80 m) d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm EVALUATE: It was important to recognize that the total displacement was h + d ; gravity continues to do work as the book moves against the spring. Also note that with the spring compressed 0.12 m it exerts an upward force (192 N) greater than the weight of the book (11.8 N). The book will be accelerated upward from this position. IDENTIFY: Use energy methods. There are changes in both elastic and gravitational potential energy. SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.20. The spring force and gravity are the only forces doing work on the cheese, so Wother = 0 and U = U grav + U el .

Figure 7.20 EXECUTE: Cheese released from rest implies K1 = 0.

At the maximum height v2 = 0 so K 2 = 0. U1 = U1,el + U1,grav y1 = 0 implies U1,grav = 0 2 U1,el = 1 kx1 = 1 (1800 N/m)(0.15 m) 2 = 20.25 J 2 2

(Here x1 refers to the amount the spring is stretched or compressed when the cheese is at position 1; it is not the x-coordinate of the cheese in the coordinate system shown in the sketch.) U 2 = U 2,el + U 2,grav U 2,grav = mgy2 , where y2 is the height we are solving for. U 2,el = 0 since now the spring is no longer compressed. Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el = U 2,grav 20.25 J 20.25 J = = 1.72 m mg (1.20 kg)(9.80 m/s 2 ) EVALUATE: The description in terms of energy is very simple; the elastic potential energy originally stored in the spring is converted into gravitational potential energy of the system. IDENTIFY: Apply Eq. (7.13). SET UP: Wother = 0. As in Example 7.7, K1 = 0 and U1 = 0.0250 J. y2 =

7.21.

2(0.0210J) = ±0.092m. 5.00N/m The glider has this speed when the spring is stretched 0.092 m or compressed 0.092 m. EVALUATE: Example 7.7 showed that vx = 0.30 m/s when x = 0.0800 m. As x increases, vx decreases, EXECUTE: For v2 = 0.20 m/s, K 2 = 0.0040 J. U 2 = 0.0210 J = 1 kx 2 , and x = ± 2

so our result of vx = 0.20 m/s at x = 0.092 m is consistent with the result in the example.

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7-8 7.22.

Chapter 7 IDENTIFY and SET UP: Use energy methods. The elastic potential energy changes. In part (a) solve for K 2

and from this obtain v2 . In part (b) solve for U1 and from this obtain x1. (a) K1 + U1 + Wother = K 2 + U 2

point 1: the glider is at its initial position, where x1 = 0.100 m and v1 = 0 point 2: the glider is at x = 0 2 EXECUTE: K1 = 0 (released from rest), K 2 = 1 mv2 2

2 U1 = 1 kx1 , U 2 = 0, Wother = 0 (only the spring force does work) 2

Thus

1 kx 2 2 1

2 = 1 mv2 . (The initial potential energy of the stretched spring is converted entirely into kinetic 2

energy of the glider.)

v2 = x1

k 5.00 n/m = (0.100 m) = 0.500 m/s m 0.200 kg
1 kx 2 2 1 2 = 1 mv2 2

(b) The maximum speed occurs at x = 0, so the same equation applies.

0.200kg m = 2.50 m/s = 0.500 m 5.00N/m k EVALUATE: Elastic potential energy is converted into kinetic energy. A larger x1 gives a larger v2 . x1 = v2 7.23. IDENTIFY: Only the spring does work and Eq. (7.11) applies. a =

F − kx , where F is the force the = m m

spring exerts on the mass. SET UP: Let point 1 be the initial position of the mass against the compressed spring, so K1 = 0 and U1 = 11.5 J. Let point 2 be where the mass leaves the spring, so U el,2 = 0. EXECUTE: (a) K1 + U el,1 = K 2 + U el,2 gives U el,1 = K 2 . v2 = 2U el,1 m = 2(11.5 J) = 3.03 m/s. 2.50 kg

1 mv 2 2 2

= U el,1 and

K is largest when U el is least and this is when the mass leaves the spring. The mass achieves its maximum speed of 3.03 m/s as it leaves the spring and then slides along the surface with constant speed. (b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has 2U el 2(11.5 J) =2 = −0.0959 m. The minus sign its maximum compression. U el = 1 kx 2 so x = − 2 2500 N/m k kx (2500 N/m)(−0.0959 m) = 95.9 m/s 2 . indicates compression. F = − kx = max and ax = − = − m 2.50 kg EVALUATE: If the end of the spring is displaced to the left when the spring is compressed, then ax in part 7.24.

(b) is to the right, and vice versa. (a) IDENTIFY and SET UP: Use energy methods. Both elastic and gravitational potential energy changes. Work is done by friction. Choose point 1 as in Example 7.9 and let that be the origin, so y1 = 0. Let point 2 be 1.00 m below point 1, so y2 = −1.00 m. EXECUTE: K1 + U1 + Wother = K 2 + U 2

2 K1 = 1 mv1 = 1 (2000 kg)(25 m/s)2 = 625,000 J, U1 = 0 2 2

Wother = − f y2 = −(17 ,000 N)(1.00 m) = −17 ,000 J
2 K 2 = 1 mg 2 2

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Potential Energy and Energy Conservation
2 U 2 = U 2,grav + U 2,el = mgy2 + 1 ky2 2

7-9

U 2 = (2000 kg)(9.80 m/s 2 )(−1.00 m) + 1 (1.41× 105 N/m)(1.00 m)2 2 U 2 = −19,600 J + 70,500 J = +50,900 J 2 Thus 625,000 J − 17,000 J = 1 mv2 + 50,900 J 2

1 mv 2 2 2

= 557,100 J
2(557,100 J) = 23.6 m/s 2000 kg

v2 =

EVALUATE: The elevator stops after descending 3.00 m. After descending 1.00 m it is still moving but has slowed down. (b) IDENTIFY: Apply ΣF = ma to the elevator. We know the forces and can solve for a . SET UP: The free-body diagram for the elevator is given in Figure 7.24. EXECUTE: Fspr = kd , where d is the distance

the spring is compressed ΣFy = ma y f k + Fspr − mg = ma f k + kd − mg = ma Figure 7.24

7.25.

f k + kd − mg 17,000 N + (1.41 × 105 N/m)(1.00 m) − (2000 kg)(9.80 m/s 2 ) = = 69.2 m/s 2 2000 kg m We calculate that a is positive, so the acceleration is upward. EVALUATE: The velocity is downward and the acceleration is upward, so the elevator is slowing down at this point. Note that a = 7.1g ; this is unacceptably high for an elevator. IDENTIFY: Apply Eq. (7.13) and F = ma. SET UP: Wother = 0. There is no change in U grav . K1 = 0, U 2 = 0. a= EXECUTE:

1 kx 2 2 2 2 = 1 mvx . The relations for m, vx , k and x are kx 2 = mvx and kx = 5mg . 2 2 vx , and substituting this into the second gives 5g

Dividing the first equation by the second gives x = k = 25 mg 2 2 vx

. = 4.46 × 105 N/m

(a) k = 25 (b) x =

(1160 kg)(9.80 m/s 2 )2 (2.50 m/s) 2 = 0.128 m

(2.50 m/s) 2 5(9.80 m/s 2 )

EVALUATE: Our results for k and x do give the required values for ax and vx :

7.26.

kx (4.46 × 105 N/m)(0.128 m) k = 2.5 m/s. = = 49.2 m/s 2 = 5.0 g and vx = x m 1160 kg m IDENTIFY: The spring force is conservative but the force of friction is nonconservative. Energy is conserved during the process. Initially all the energy is stored in the spring, but part of this goes to kinetic energy, part remains as elastic potential energy, and the rest does work against friction. SET UP: Energy conservation: K1 + U1 + Wother = K 2 + U 2 , the elastic energy in the spring is U = 1 kx 2 , 2 ax = and the work done by friction is Wf = − f k s = − μ k mgs.

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7-10

Chapter 7 EXECUTE: The initial and final elastic potential energies are 2 2 U1 = 1 kx1 = 1 (840 N/m)(0.0300 m) 2 = 0.378 J and U 2 = 1 kx2 = 1 (840 N/m)(0.0100 m)2 = 0.0420 J. The 2 2 2 2 2 initial and final kinetic energies are K1 = 0 and K 2 = 1 mv2 . The work done by friction is 2

Wother = W f k = − f k s = − μk mgs = −(0.40)(2.50 kg)(9.8 m/s 2 )(0.0200 m) = −0.196 J. Energy conservation 2 gives K 2 = 1 mv2 = K1 + U1 + Wother − U 2 = 0.378 J + (−0.196 J) − 0.0420 J = 0.140 J. Solving for v2 gives 2

v2 =
7.27.

2K2 2(0.140 J) = = 0.335 m/s. 2.50 kg m

EVALUATE: Mechanical energy is not conserved due to friction. IDENTIFY: Apply W f k = f k s cos φ . f k = μ k n. SET UP: For a circular trip the distance traveled is d = 2π r. At each point in the motion the friction force and the displacement are in opposite directions and φ = 180°. Therefore, W f k = − f k d = − f k (2π r ). n = mg

so f k = μ k mg.
EXECUTE: (a) W f k = − μk mg 2π r = −(0.250)(10.0 kg)(9.80 m/s2 )(2π )(2.00 m) = −308 J. (b) The distance along the path doubles so the work done doubles and becomes −616 J. (c) The work done for a round trip displacement is not zero and friction is a nonconservative force. EVALUATE: The direction of the friction force depends on the direction of motion of the object and that is why friction is a nonconservative force. IDENTIFY: Wgrav = mg cos φ . SET UP: When he moves upward, φ = 180° and when he moves downward, φ = 0°. When he moves parallel to the ground, φ = 90°. EXECUTE: (a) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos180° = −5100 J. (b) Wgrav = (75 kg)(9.80 m/s 2 )(7.0 m)cos0° = +5100 J. (c) φ = 90° in each case and Wgrav = 0 in each case. (d) The total work done on him by gravity during the round trip is −5100 J + 5100 J = 0. (e) Gravity is a conservative force since the total work done for a round trip is zero. EVALUATE: The gravity force is independent of the position and motion of the object. When the object moves upward gravity does negative work and when the object moves downward gravity does positive work. IDENTIFY: Since the force is constant, use W = Fs cosφ . SET UP: For both displacements, the direction of the friction force is opposite to the displacement and φ = 180°. EXECUTE: (a) When the book moves to the left, the friction force is to the right, and the work is −(1.2 N)(3.0 m) = −3.6 J. (b) The friction force is now to the left, and the work is again −3.6 J. (c) −7.2 J. (d) The net work done by friction for the round trip is not zero, and friction is not a conservative force. EVALUATE: The direction of the friction force depends on the motion of the object. For the gravity force, which is conservative, the force does not depend on the motion of the object. IDENTIFY and SET UP: The force is not constant so we must use Eq. (6.14) to calculate W. The properties of work done by a conservative force are described in Section 7.3.

7.28.

7.29.

7.30.

ˆ W = ∫ F ⋅ dl , F = −α x 2 i
1

2

EXECUTE: (a) dl = dyˆ (x is constant; the displacement is in the + y -direction ) j ˆ j F ⋅ dl = 0 (since i ⋅ ˆ = 0) and thus W = 0.

ˆ (b) dl = dxi
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Potential Energy and Energy Conservation

7-11

ˆ ˆ F ⋅ dl = (−α x 2i ) ⋅ (dxi ) = −α x 2 dx
3 3 W = ∫ ( −α x 2 ) dx = − 1 ax3 |x2 = − 1 α ( x2 − x1 ) = − x 3 3

x2

x1

1

12 N/m 2 ((0.300 m)3 − (0.10 m)3 ) = −0.10 J 3

ˆ (c) dl = dxi as in part (b), but now x1 = 0.30 m and x2 = 0.10 m 3 3 W = − 1 α ( x2 − x1 ) = +0.10 J 3

(d) EVALUATE: The total work for the displacement along the x-axis from 0.10 m to 0.30 m and then back to 0.10 m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when the starting and ending points are the same, so the force is conservative. 3 3 3 3 EXECUTE: Wx1 → x2 = − 1 α ( x2 − x1 ) = 1 α x1 − 1 α x2 3 3 3

The definition of the potential energy function is Wx1 → x2 = U1 − U 2 . Comparison of the two expressions for W gives U = 1 α x3. This does correspond to U = 0 when x = 0. 3 EVALUATE: In part (a) the work done is zero because the force and displacement are perpendicular. In part (b) the force is directed opposite to the displacement and the work done is negative. In part (c) the force and displacement are in the same direction and the work done is positive. IDENTIFY and SET UP: The friction force is constant during each displacement and Eq. (6.2) can be used to calculate work, but the direction of the friction force can be different for different displacements. f = μk mg = (0.25)(1.5 kg)(9.80 m/s 2 ) = 3.675 N; direction of f is opposite to the motion. EXECUTE: (a) The path of the book is sketched in Figure 7.31a.

7.31.

Figure 7.31a

For the motion from you to Beth the friction force is directed opposite to the displacement s and W1 = − fs = −(3.675 N)(8.0 m) = −29.4 J. For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and W2 = −29.4 J. Wtot = W1 + W2 = −29.4 J − 29.4 J = −59 J (b) The path of the book is sketched in Figure 7.31b.

s = 2(8.0 m)2 = 11.3 m

Figure 7.31b

f is opposite to s , so W = − fs = −(3.675 N)(11.3 m) = −42 J

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7-12

Chapter 7 (c)

For the motion from you to Kim (Figure 7.31c) W = − fs W = −(3.675 N)(8.0 m) = −29.4 J

Figure 7.31c

For the motion from Kim to you (Figure 7.31d) W = − fs = −29.4 J

Figure 7.31d

7.32.

The total work for the round trip is −29.4 J − 29.4 J = −59 J. (d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero. Both these results show that the friction force is nonconservative. IDENTIFY: Some of the initial gravitational potential energy is converted to kinetic energy, but some of it is lost due to work by the nonconservative friction force. SET UP: The energy of the box at the edge of the roof is given by: Emech, f = Emech, i − f k s. Setting yf = 0 at this point, yi = (4.25 m) sin36° = 2.50 m. Furthermore, by substituting Ki = 0 and K f = 1 mvf 2 2 into the conservation equation, 1 mv 2 f 2

= mgyi − f k s or vf = 2 gyi − 2 f k sg/w = 2 g ( yi − f k s/w).

EXECUTE: vf = 2(9.80 m/s 2 ) [ (2.50 m) − (22.0 N)(4.25 m)/(85.0 N) ] = 5.24 m/s. EVALUATE: Friction does negative work and removes mechanical energy from the system. In the absence of friction the final speed of the toolbox would be 7.00 m/s. IDENTIFY: Some of the mechanical energy of the skier is converted to internal energy by the nonconservative force of friction on the rough patch. SET UP: For part (a) use Emech, f = Emech, i − f k s where f k = μk mg . Let yf = 0 at the bottom of the hill;

7.33.

then yi = 2.50 m along the rough patch. The energy equation is thus as the negative of the work done by friction: −W f = + f k s = + μk mgs.

1 mv 2 f 2

= 1 mvi 2 + mgyi − μ k mgs. 2

Solving for her final speed gives vf = vi 2 + 2 gyi − 2 μ k gs . For part (b), the internal energy is calculated EXECUTE: (a) vf = (6.50 m/s) 2 + 2(9.80 m/s 2 )(2.50 m) − 2(0.300)(9.80 m/s 2 )(3.50 m) = 8.41 m/s. (b) Internal energy = μk mgs = (0.300)(62.0 kg)(9.80 m/s 2 )(3.50 m) = 638 J. EVALUATE: Without friction the skier would be moving faster at the bottom of the hill than at the top, but in this case she is moving slower because friction converted some of her initial kinetic energy into internal energy. IDENTIFY and SET UP: Use Eq. (7.17) to calculate the force from U ( x ). Use coordinates where the origin

7.34.

is at one atom. The other atom then has coordinate x. EXECUTE: dU d ⎛ C6 ⎞ d ⎛ 1 ⎞ 6C = − ⎜ − 6 ⎟ = +C6 ⎜ 6 ⎟ = − 76 Fx = − dx dx ⎝ x ⎠ dx ⎝ x ⎠ x The minus sign mean that Fx is directed in the − x-direction, toward the origin. The force has magnitude 6C6 /x 7 and is attractive. EVALUATE: U depends only on x so F is along the x-axis; it has no y or z components.

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Potential Energy and Energy Conservation 7.35. IDENTIFY: Apply Eq. (7.16). SET UP: The sign of Fx indicates its direction.

7-13

dU = −4α x3 = −(4.8 J/m 4 ) x3. Fx (−0.800 m) = −(4.8 J/m 4 )( −0.80 m)3 = 2.46 N. The dx force is in the + x -direction. EVALUATE: Fx > 0 when x < 0 and Fx < 0 when x > 0, so the force is always directed towards the

EXECUTE: Fx = −

7.36.

origin. IDENTIFY: Apply Eq. (7.18). d ⎛ 1 ⎞ 2 d ⎛ 1 ⎞ 2 SET UP: ⎜ ⎟=− 3. ⎜ 2 ⎟ = − 3 and dy ⎜ y 2 ⎟ dx ⎝ x ⎠ y x ⎝ ⎠ ∂U ˆ ∂U ˆ α i− j since U has no z-dependence. ∂U = −2α and ∂U = −23 , so EXECUTE: F = − ∂x ∂y x3 y ∂x ∂y ⎛ −2 ˆ −2 ⎞ ⎛ i j ⎞ F = −α ⎜ 3 i + 3 ˆ ⎟ = 2α ⎜ 3 + 3 ⎟ . j⎟ ⎜x ⎜x y ⎠ y ⎟ ⎝ ⎝ ⎠ EVALUATE: Fx and x have the same sign and Fy and y have the same sign. When x > 0, Fx is in the

7.37.

+ x-direction, and so forth. IDENTIFY: From the potential energy function of the block, we can find the force on it, and from the force we can use Newton’s second law to find its acceleration. ∂U ∂U SET UP: The force components are Fx = − and Fy = − . The acceleration components are ∂x ∂y 2 ax = Fx /m and a y = Fy /m. The magnitude of the acceleration is a = a x + a 2 and we can find its angle y

with the +x axis using tan θ = a y /a x .
EXECUTE: Fx = −

∂U ∂U = −(11.6 J/m 2 ) x and Fy = − = (10.8 J/m3 ) y 2 . At the point ∂x ∂y Fy Fx = −87.0 m/s 2 and a y = = 97.2 m/s 2 , giving m m

( x = 0.300 m, y = 0.600 m ), Fx = −(11.6 J/m 2 )(0.300 m) = −3.48 N and Fy = (10.8 J/m3 )(0.600 m) 2 = 3.89 N. Therefore a x = 2 a = a x + a 2 = 130 m/s 2 and tan θ = y

7.38.

the + x-axis. EVALUATE: The force is not constant, so the acceleration will not be the same at other points. IDENTIFY: Apply Eq. (7.16). dU SET UP: is the slope of the U versus x graph. dx dU and so the force is zero when the EXECUTE: (a) Considering only forces in the x-direction, Fx = − dx slope of the U vs x graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. EVALUATE: At point b, Fx is negative when the marble is displaced slightly to the right and Fx is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable.

97.2 , so θ = 48.2°. The direction is 132o counterclockwise from 87.0

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7-14 7.39.

Chapter 7 IDENTIFY and SET UP: Use Eq. (7.17) to calculate the force from U. At equilibrium F = 0. (a) EXECUTE: The graphs are sketched in Figure 7.39.

U=

a b − r12 r 6 dU 12a 6b F =− = + 13 − 7 dr r r

Figure 7.39 (b) At equilibrium F = 0, so

dU =0 dr F = 0 implies +12a r
13



6b r7

=0

6br 6 = 12a; solution is the equilibrium distance r0 = (2a/b)1/ 6 U is a minimum at this r; the equilibrium is stable. (c) At r = (2a/b)1/6 , U = a/r12 − b/r 6 = a (b/2a ) 2 − b(b/2a ) = −b 2 /4a. At r → ∞, U = 0. The energy that must be added is −ΔU = b 2 /4a. (d) r0 = (2a/b)1/6 = 1.13 × 10−10 m gives that

2a/b = 2.082 × 10−60 m 6 and b/4a = 2.402 × 1059 m −6 b 2 /4a = b(b/4a ) = 1.54 × 10−18 J b(2.402 × 1059 m −6 ) = 1.54 × 10−18 J and b = 6.41 × 10−78 J ⋅ m 6 . Then 2a/b = 2.082 × 10−60 m 6 gives a = (b/2)(2.082 × 10−60 m 6 ) = 1 (6.41 × 10−78 2

J ⋅ m6 ) (2.082 × 10−60 m6 ) = 6.67 × 10−138 J ⋅ m12

7.40.

EVALUATE: As the graphs in part (a) show, F (r ) is the slope of U (r ) at each r. U ( r ) has a minimum where F = 0. IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq. (7.5). SET UP: Call the blocks A and B, where A is the more massive one. v A1 = vB1 = 0. Let y = 0 for each

block to be at the initial height of that block, so y A1 = yB1 = 0. y A2 = −1.20 m and y B 2 = +1.20 m. v A2 = vB 2 = v2 = 3.00 m/s. 2 EXECUTE: Eq. (7.5) gives 0 = 1 (m A + mB )v2 + g (1.20 m)(mB − mA ) ⋅ m A + mB = 15.0 kg ⋅ 2 1 (15.0 2

kg)(3.00 m/s) 2 + (9.80 m/s 2 )(1.20 m)(15.0 kg − 2m A ). Solving for m A gives m A = 10.4 kg.

And then mB = 4.6 kg.
EVALUATE: The final kinetic energy of the two blocks is 68 J. The potential energy of block A decreases by 122 J. The potential energy of block B increases by 54 J. The total decrease in potential energy is 122 J − 54 J = 68 J, and this equals the increase in kinetic energy of the system. 7.41. IDENTIFY: Apply ΣF = ma to the bag and to the box. Apply Eq. (7.7) to the motion of the system of the box and bucket after the bag is removed. SET UP: Let y = 0 at the final height of the bucket, so y1 = 2.00 m and y2 = 0 . K1 = 0. The box and the

bucket move with the same speed v, so K 2 = 1 ( mbox + mbucket )v 2 . Wother = − f k d , with d = 2.00 m and 2 f k = μ k mbox g . Before the bag is removed, the maximum possible friction force the roof can exert on the box is (0.700)(80.0 kg + 50.0 kg)(9.80 m/s 2 ) = 892 N. This is larger than the weight of the bucket (637 N), so before the bag is removed the system is at rest.

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Potential Energy and Energy Conservation

7-15

EXECUTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag for friction to oppose. The static friction force on the box equals the weight of the bucket, 637 N. 2 (b) Eq. (7.7) gives mbucket gy1 − f k d = 1 mtot v 2 , with mtot = 145.0 kg. v = (mbucket gy1 − μ k mbox gd ). 2 mtot

v=

2 [(65.0 kg)(9.80 m/s 2 )(2.00 m) − (0.400)(80.0 kg)(9.80 m/s 2 )(2.00 m)]. 145.0 kg

7.42.

v = 2.99 m/s. EVALUATE: If we apply ΣF = ma to the box and to the bucket we can calculate their common acceleration a. Then a constant acceleration equation applied to either object gives v = 2.99 m/s, in agreement with our result obtained using energy methods. IDENTIFY: Apply Eq. (7.14). SET UP: Only the spring force and gravity do work, so Wother = 0. Let y = 0 at the horizontal surface. EXECUTE: (a) Equating the potential energy stored in the spring to the block's kinetic energy, 1 kx 2 = 1 mv 2 , or v = k x = 400 N/m (0.220 m) = 3.11 m/s. 2 2 m 2.00 kg (b) Using energy methods directly, the initial potential energy of the spring equals the final gravitational 1 kx 2 2 1 kx 2 2 1 (400 2

7.43.

= 0.821 m. (2.00 kg)(9.80 m/s 2 )sin 37.0° EVALUATE: The total energy of the system is constant. Initially it is all elastic potential energy stored in the spring, then it is all kinetic energy and finally it is all gravitational potential energy. IDENTIFY: Use the work-energy theorem, Eq. (7.7). The target variable μ k will be a factor in the work potential energy, mg sin θ done by friction. SET UP: Let point 1 be where the block is released and let point 2 be where the block stops, as shown in Figure 7.43. K1 + U1 + Wother = K 2 + U 2 Work is done on the block by the spring and by friction, so Wother = W f and U = U el .

= mgL sin θ , or L =

=

N/m)(0.220 m) 2

Figure 7.43 EXECUTE: K1 = K 2 = 0
2 U1 = U1,el = 1 kx1 = 1 (100 N/m)(0.200 m)2 = 2.00 J 2 2

U 2 = U 2 ,el = 0, since after the block leaves the spring has given up all its stored energy

Wother = W f = ( f k cos φ ) s = μk mg ( cos φ ) s = − μk mgs, since φ = 180° (The friction force is directed opposite to the displacement and does negative work.) Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1,el + W f = 0

μ k mgs = U1,el

μk =

U1,el mgs

=

2.00 J (0.50 kg)(9.80 m/s 2 )(1.00 m)

= 0.41.

EVALUATE: U1,el + W f = 0 says that the potential energy originally stored in the spring is taken out of the

system by the negative work done by friction.

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7-16 7.44.

Chapter 7 IDENTIFY: Apply Eq. (7.14). Calculate f k from the fact that the crate slides a distance x = 5.60 m before coming to rest. Then apply Eq. (7.14) again, with x = 2.00 m. SET UP: U1 = U el = 360 J. U 2 = 0. K1 = 0. Wother = − f k x. EXECUTE: Work done by friction against the crate brings it to a halt: U1 = −Wother .

360 J = 64.29 N. 5.60 m The friction force working over a 2.00-m distance does work equal to − f k x = −(64.29 N)(2.00 m) = −128.6 J. The kinetic energy of the crate at this point is thus f k x = potential energy of compressed spring , and f k =

360 J − 128.6 J = 231.4 J, and its speed is found from mv 2 /2 = 231.4 J, so v =

2(231.4 J) = 3.04 m/s. 50.0 kg

7.45.

EVALUATE: The energy of the compressed spring goes partly into kinetic energy of the crate and is partly removed by the negative work done by friction. After the crate leaves the spring the crate slows down as friction does negative work on it. IDENTIFY: The mechanical energy of the roller coaster is conserved since there is no friction with the track. We must also apply Newton’s second law for the circular motion. SET UP: For part (a), apply conservation of energy to the motion from point A to point B: K B + U grav , B = K A + U grav ,A with K A = 0. Defining y B = 0 and y A = 13.0 m, conservation of energy

becomes

1 mvB 2 2

= mgy A or vB = 2 gy A . In part (b), the free-body diagram for the roller coaster car at

point B is shown in Figure 7.45. ΣFy = ma y gives mg + n = marad , where arad = v 2 /r. Solving for the

⎛ v2 ⎞ normal force gives n = m ⎜ − g ⎟ . ⎜ r ⎟ ⎝ ⎠

Figure 7.45 EXECUTE: (a) vB = 2(9.80 m/s 2 )(13.0 m) = 16.0 m/s.

7.46.

⎡ (16.0 m/s) 2 ⎤ (b) n = (350 kg) ⎢ − 9.80 m/s 2 ⎥ = 1.15 × 104 N. ⎢ 6.0 m ⎥ ⎣ ⎦ EVALUATE: The normal force n is the force that the tracks exert on the roller coaster car. The car exerts a force of equal magnitude and opposite direction on the tracks. IDENTIFY: Apply Eq. (7.14) to relate h and vB . Apply ΣF = ma at point B to find the minimum speed required at B for the car not to fall off the track. 2 2 SET UP: At B, a = vB /R , downward. The minimum speed is when n → 0 and mg = mvB /R. The minimum speed required is vB = gR . K1 = 0 and Wother = 0. 2 EXECUTE: (a) Eq. (7.14) applied to points A and B gives U A − U B = 1 mvB . The speed at the top must be 2

at least

1 5 gR . Thus, mg ( h − 2 R) > mgR, or h > R. 2 2 (b) Apply Eq. (7.14) to points A and C. U A − U C = (2.50) Rmg = KC , so

vC = (5.00) gR = (5.00)(9.80 m/s 2 )(20.0 m) = 31.3 m/s.
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Potential Energy and Energy Conservation
2 vC = 49.0 m/s 2 . The tangential direction is down, the normal force at R

7-17

The radial acceleration is arad =

point C is horizontal, there is no friction, so the only downward force is gravity, and atan = g = 9.80 m/s 2 . EVALUATE: If h > 5 R, then the downward acceleration at B due to the circular motion is greater than g 2

and the track must exert a downward normal force n. n increases as h increases and hence vB increases. 7.47. (a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom. SET UP: Let point 1 be where the piece of wood is released and point 2 be just before it enters the rough bottom. Let y = 0 be at point 2. EXECUTE: U1 = K 2 gives K 2 = mgy1 = 78.4 J. IDENTIFY: Now apply work-energy relation to the motion along the rough bottom. SET UP: Let point 1 be where it enters the rough bottom and point 2 be where it stops.

K1 + U1 + Wother = K 2 + U 2
EXECUTE: Wother = W f = − μ k mgs, K 2 = U1 = U 2 = 0; K1 = 78.4 J

78.4 J − μk mgs = 0; solving for s gives s = 20.0 m. The wood stops after traveling 20.0 m along the rough bottom. (b) Friction does −78.4 J of work. EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system. IDENTIFY: Apply Eq. (7.14) to the rock. Wother = W f . k

7.48.

SET UP: Let y = 0 at the foot of the hill, so U1 = 0 and U 2 = mgh, where h is the vertical height of the

rock above the foot of the hill when it stops. EXECUTE: (a) At the maximum height, K 2 = 0. Eq. (7.14) gives K Bottom + W f = U Top . k

h 1 2 1 2 mv − μ k mg cosθ d = mgh. d = h / sin θ , so v0 − μ k g cosθ = gh. 2 0 2 sin θ 1 cos 40° (15 m/s)2 − (0.20)(9.8 m/s 2 ) h = (9.8 m/s 2 )h and h = 9.3 m. 2 sin 40° (b) Compare maximum static friction force to the weight component down the plane. fs = μs mg cosθ = (0.75)(28 kg)(9.8 m/s 2 )cos 40° = 158 N. mg sinθ = (28 kg)(9.8 m/s 2 )(sin 40°) = 176 N > fs , so the rock will slide down.

(c) Use same procedure as in part (a), with h = 9.3 m and v B being the speed at the bottom of the hill.

U Top + W f = K B . mgh − μ k mg cosθ
k

h 1 = mv 2 and sin θ 2 B

v B = 2 gh − 2 μk gh cosθ / sin θ = 11.8 m/s.
EVALUATE: For the round trip up the hill and back down, there is negative work done by friction and the speed of the rock when it returns to the bottom of the hill is less than the speed it had when it started up the hill. IDENTIFY: Apply Eq. (7.7) to the motion of the stone. SET UP: K1 + U1 + Wother = K 2 + U 2 Let point 1 be point A and point 2 be point B. Take y = 0 at point B. 2 2 EXECUTE: mgy1 + 1 mv1 = 1 mv2 , with h = 20.0 m and v1 = 10.0 m/s 2 2 2 v2 = v1 + 2 gh = 22.2 m/s

7.49.

EVALUATE: The loss of gravitational potential energy equals the gain of kinetic energy. (b) IDENTIFY: Apply Eq. (7.8) to the motion of the stone from point B to where it comes to rest against the spring. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7-18

Chapter 7 SET UP: Use K1 + U1 + Wother = K 2 + U 2 , with point 1 at B and point 2 where the spring has its maximum

compression x. 2 EXECUTE: U1 = U 2 = K 2 = 0; K1 = 1 mv1 with v1 = 22.2 m/s 2 Wother = W f + Wel = -μ k mgs − 1 kx 2 , with s = 100 m + x 2 The work-energy relation gives K1 + Wother = 0. 1 mv 2 1 2

− μ k mgs − 1 kx 2 = 0 2

Putting in the numerical values gives x 2 + 29.4 x − 750 = 0. The positive root to this equation is x = 16.4 m. EVALUATE: Part of the initial mechanical (kinetic) energy is removed by friction work and the rest goes into the potential energy stored in the spring. (c) IDENTIFY and SET UP: Consider the forces. EXECUTE: When the spring is compressed x = 16.4 m the force it exerts on the stone is Fel = kx = 32.8 N. The maximum possible static friction force is max fs = μs mg = (0.80)(15.0 kg)(9.80 m/s 2 ) = 118 N.

7.50.

EVALUATE: The spring force is less than the maximum possible static friction force so the stone remains at rest. IDENTIFY: Once the block leaves the top of the hill it moves in projectile motion. Use Eq. (7.14) to relate the speed vB at the bottom of the hill to the speed vTop at the top and the 70 m height of the hill. SET UP: For the projectile motion, take + y to be downward. a x = 0, a y = g . v0 x = vTop , v0 y = 0. For

the motion up the hill only gravity does work. Take y = 0 at the base of the hill. EXECUTE: First get speed at the top of the hill for the block to clear the pit. y =

1 2 gt . 2

1 40 m 20 m = (9.8 m/s 2 )t 2 . t = 2.0 s. Then vTopt = 40 m gives vTop = = 20 m/s. 2 2.0 s Energy conservation applied to the motion up the hill: K Bottom = U Top + K Top gives 1 2 1 2 2 mvB = mgh + mvTop . vB = vTop + 2 gh = (20 m/s) 2 + 2(9.8 m/s 2 )(70 m) = 42 m/s. 2 2 EVALUATE: The result does not depend on the mass of the block. IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the person. SET UP: Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at point 2. y1 = 41.0 m. Wother = - 1 kx 2 , where x = 11.0 m is the amount the cord is stretched at 2

7.51.

point 2. The cord does negative work. EXECUTE: K1 = K 2 = U 2 = 0, so mgy1 − 1 kx 2 = 0 and k = 631 N/m. 2 Now apply F = kx to the test pulls: F = kx so x = F /k = 0.602 m. EVALUATE: All his initial gravitational potential energy is taken away by the negative work done by the force exerted by the cord, and this amount of energy is stored as elastic potential energy in the stretched cord. IDENTIFY: Apply Eq. (7.14) to the motion of the skier from the gate to the bottom of the ramp. SET UP: Wother = -4000 J. Let y = 0 at the bottom of the ramp. EXECUTE: For the skier to be moving at no more than 30.0 m/s, his kinetic energy at the bottom of the mv 2 (85.0 kg)(30.0 m/s) 2 = = 38,250 J. Friction does −4000 J of work on 2 2 him during his run, which means his combined U and K at the top of the ramp must be no more than mv 2 (85.0 kg)(2.0 m/s) 2 38,250 J + 4000 J = 42,250 J. His K at the top is = = 170 J. His U at the top 2 2 ramp can be no bigger than

7.52.

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Potential Energy and Energy Conservation

7-19

should thus be no more than 42,250 J − 170 J = 42,080 J, which gives a height above the bottom of the 42,080 J 42,080 J ramp of h = = = 50.5 m. mg (85.0 kg)(9.80 m/s 2 ) EVALUATE: In the absence of air resistance, for this h his speed at the bottom of the ramp would be 31.5 m/s. The work done by air resistance is small compared to the kinetic and potential energies that enter into the calculation. IDENTIFY: Use the work-energy theorem, Eq. (7.7). Solve for K 2 and then for v2 . SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel, as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by the 4400 N force. K1 + U1 + Wother = K 2 + U 2

7.53.

Take y = 0 at his initial position.
2 EXECUTE: K1 = 0, K 2 = 1 mv2 2

Wother = Wfric = − fs Wother = −(40 N)(4.0 m) = −160 J Figure 7.53

U1,grav = 0, U1,el = 1 kd 2 , where d is the distance the spring is initially compressed. 2 F = kd so d = F 4400 N = = 4.00 m k 1100 N/m

and U1,el = 1 (1100 N/m)(4.00 m) 2 = 8800 J 2 U 2 ,grav = mgy2 = (60 kg)(9.80 m/s 2 )(2.5 m) = 1470 J, U 2 ,el = 0 Then K1 + U1 + Wother = K 2 + U 2 gives 2 8800 J − 160 J = 1 mv2 + 1470 J 2
1 mv 2 2 2

= 7170 J and v2 =

2(7170 J) = 15.5 m/s 60 kg

7.54.

EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy. IDENTIFY: To be at equilibrium at the bottom, with the spring compressed a distance x0 , the spring force

must balance the component of the weight down the ramp plus the largest value of the static friction, or kx0 = w sin θ + f . Apply Eq. (7.14) to the motion down the ramp. SET UP: K 2 = 0, K1 = 1 mv 2 , where v is the speed at the top of the ramp. Let U 2 = 0, so U1 = wL sin θ , 2

where L is the total length traveled down the ramp. 1 2 1 EXECUTE: Eq. (7.14) gives kx0 = ( w sin θ − f ) L + mv 2 . With the given parameters, 2 2 kx0 = 1.10 × 103 N. Solving for k gives k = 2440 N/m.

1 kx 2 2 0

= 248 J and

EVALUATE: x0 = 0.451 m. w sin θ = 551 N. The decrease in gravitational potential energy is only slightly

larger than the amount of mechanical energy removed by the negative work done by friction. 1 mv 2 = 243 J. The energy stored in the spring is only slightly larger than the initial kinetic energy of the 2 crate at the top of the ramp.

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7-20 7.55.

Chapter 7 IDENTIFY: Apply Eq. (7.7) to the system consisting of the two buckets. If we ignore the inertia of the pulley we ignore the kinetic energy it has. SET UP: K1 + U1 + Wother = K 2 + U 2 . Points 1 and 2 in the motion are sketched in Figure 7.55.

Figure 7.55

The tension force does positive work on the 4.0 kg bucket and an equal amount of negative work on the 12.0 kg bucket, so the net work done by the tension is zero. Work is done on the system only by gravity, so Wother = 0 and U = U grav EXECUTE: K1 = 0

2 K 2 = 1 m Av 2 ,2 + 1 mB vB ,2 But since the two buckets are connected by a rope they move together and have A 2 2

the same speed: v A,2 = vB ,2 = v2 .
2 2 Thus K 2 = 1 ( mA + mB )v2 = (8.00 kg)v2 . 2

U1 = m A gy A,1 = (12.0 kg)(9.80 m/s 2 )(2.00 m) = 235.2 J. U 2 = mB gyB ,2 = (4.0 kg)(9.80 m/s 2 )(2.00 m) = 78.4 J. Putting all this into K1 + U1 + Wother = K 2 + U 2 gives U1 = K 2 + U 2 2 235.2 J = (8.00 kg)v2 + 78.4 J

v2 =

235.2 J − 78.4 J = 4.4 m/s 8.00 kg

EVALUATE: The gravitational potential energy decreases and the kinetic energy increases by the same amount. We could apply Eq. (7.7) to one bucket, but then we would have to include in Wother the work 7.56.

done on the bucket by the tension T. IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 to the motion of the rocket from the starting point to the base of the ramp. Wother is the work done by the thrust and by friction. SET UP: Let point 1 be at the starting point and let point 2 be at the base of the ramp. v1 = 0,

v2 = 50.0 m/s. Let y = 0 at the base and take + y upward. Then y2 = 0 and y1 = d sin 53°, where d is the distance along the ramp from the base to the starting point. Friction does negative work. EXECUTE: K1 = 0, U 2 = 0. U1 + Wother = K 2 . Wother = (2000 N)d − (500 N)d = (1500 N) d . 2 mgd sin 53° + (1500 N) d = 1 mv2 . 2 2 mv2 (1500 kg)(50.0 m/s) 2 = = 142 m. 2[mg sin 53° + 1500 N] 2[(1500 kg)(9.80 m/s 2 )sin 53° + 1500 N]

d=

EVALUATE: The initial height is y1 = (142 m)sin 53° = 113 m. An object free-falling from this distance

attains a speed v = 2 gy1 = 47.1 m/s. The rocket attains a greater speed than this because the forward thrust is greater than the friction force. © Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Potential Energy and Energy Conservation 7.57. IDENTIFY: Apply K1 + U1 + Wother = K 2 + U 2 SET UP: U1 = U 2 = K 2 = 0. Wother = W f = − μk mgs, with s = 280 ft = 85.3 m EXECUTE: (a) The work-energy expression gives 1 mv 2 1 2

7-21

− μ k mgs = 0.

v1 = 2 μk gs = 22.4 m/s = 50 mph; the driver was speeding.
(b) 15 mph over speed limit so $150 ticket. EVALUATE: The negative work done by friction removes the kinetic energy of the object. IDENTIFY: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy. SET UP: Since the two animals are equidistant from the axis, they each have the same speed v. EXECUTE: One mass rises while the other falls, so the net loss of potential energy is (0.500 kg − 0.200 kg)(9.80 m/s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the

7.58.

animals and is equal to

1 m v2 , 2 tot

and v =

2(1.176 J) = 1.83 m/s. (0.700 kg)

7.59.

EVALUATE: The mouse gains both gravitational potential energy and kinetic energy. The rat’s gain in kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse. (a) IDENTIFY and SET UP: Apply Eq. (7.7) to the motion of the potato. Let point 1 be where the potato is released and point 2 be at the lowest point in its motion, as shown in Figure 7.59a. K1 + U1 + Wother = K 2 + U 2

y1 = 2.50 m y2 = 0 The tension in the string is at all points in the motion perpendicular to the displacement, so Wr = 0 The only force that does work on the potato is gravity, so Wother = 0.

Figure 7.59a
2 2 EXECUTE: K1 = 0, K 2 = 1 mv2 , U1 = mgy1, U 2 = 0. Thus U1 = K 2 . mgy1 = 1 mv2 , which gives 2 2

v2 = 2 gy1 = 2(9.80 m/s2 )(2.50 m) = 7.00 m/s.
EVALUATE: The speed v2 is the same as if the potato fell through 2.50 m. (b) IDENTIFY: Apply ΣF = ma to the potato. The potato moves in an arc of a circle so its acceleration is

arad , where arad = v 2/R and is directed toward the center of the circle. Solve for one of the forces, the tension T in the string. SET UP: The free-body diagram for the potato as it swings through its lowest point is given in Figure 7.59b. The acceleration arad is directed in toward the center of the circular path, so at this point it is upward.

Figure 7.59b

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7-22

Chapter 7

⎛ v2 ⎞ EXECUTE: ΣFy = ma y gives T − mg = marad. Solving for T gives T = m( g + arad ) = m ⎜ g + 2 ⎟ , where ⎜ R⎟ ⎝ ⎠ the radius R for the circular motion is the length L of the string. It is instructive to use the algebraic expression for v2 from part (a) rather than just putting in the numerical value: v2 = 2 gy1 = 2 gL , so ⎛ v2 ⎞ 2 gL ⎞ ⎛ 2 v2 = 2 gL. Then T = m ⎜ g + 2 ⎟ = m ⎜ g + ⎟ = 3mg . The tension at this point is three times the weight ⎜ L⎟ L ⎠ ⎝ ⎝ ⎠

of the potato, so T = 3mg = 3(0.300 kg)(9.80 m/s 2 ) = 8.82 N. 7.60. EVALUATE: The tension is greater than the weight; the acceleration is upward so the net force must be upward. IDENTIFY: Eq. (7.14) says Wother = K 2 + U 2 − ( K1 + U1 ). Wother is the work done on the baseball by the

force exerted by the air. 2 SET UP: U = mgy. K = 1 mv 2 , where v 2 = vx + v 2 . y 2 EXECUTE: (a) The change in total energy is the work done by the air, ⎛1 2 2 ⎞ Wother = ( K 2 + U 2 ) − ( K1 + U1 ) = m ⎜ (v2 − v1 ) + gy2 ⎟ . 2 ⎝ ⎠

Wother = (0.145 kg)((1/2[(18.6 m/s) 2 − (30.0 m/s)2 − (40.0 m/s)2 ] + (9.80 m/s 2 )(53.6 m)). Wother = -80.0 J. (b) Similarly, Wother = ( K3 + U 3 ) − ( K 2 + U 2 ).

Wother = (0.145 kg)((1/2)[(11.9 m/s)2 + (−28.7 m/s) 2 − (18.6 m/s) 2 ] − (9.80 m/s 2 )(53.6 m)). Wother = -31.3 J. (c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by the air is smaller in magnitude. EVALUATE: The initial kinetic energy of the baseball is 1 (0.145 kg)(50.0 m/s)2 = 181 J. For the total 2

7.61.

motion from the ground, up to the maximum height, and back down the total work done by the air is 111 J. The ball returns to the ground with 181 J − 111 J = 70 J of kinetic energy and a speed of 31 m/s, less than its initial speed of 50 m/s. IDENTIFY and SET UP: There are two situations to compare: stepping off a platform and sliding down a pole. Apply the work-energy theorem to each. (a) EXECUTE: Speed at ground if steps off platform at height h: K1 + U1 + Wother = K 2 + U 2 2 2 mgh = 1 mv2 , so v2 = 2 gh 2

Motion from top to bottom of pole: (take y = 0 at bottom) K1 + U1 + Wother = K 2 + U 2 2 mgd − fd = 1 mv2 2 2 Use v2 = 2 gh and get mgd − fd = mgh fd = mg ( d − h) f = mg (d − h) / d = mg (1 − h /d ) EVALUATE: For h = d this gives f = 0 as it should (friction has no effect).

For h = 0, v2 = 0 (no motion). The equation for f gives f = mg in this special case. When f = mg the forces on him cancel and he doesn’t accelerate down the pole, which agrees with v2 = 0. (b) EXECUTE:

f = mg (1 − h/d ) = (75 kg)(9.80 m/s 2 )(1 − 1.0 m/2.5 m) = 441 N.

(c) Take y = 0 at bottom of pole, so y1 = d and y2 = y.

K1 + U1 + Wother = K 2 + U 2

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Potential Energy and Energy Conservation

7-23

0 + mgd − f ( d − y ) = 1 mv 2 + mgy 2
1 mv 2 2

= mg (d − y ) − f (d − y )
1 mv 2 2

Using f = mg (1 − h/d ) gives
1 mv 2 2

= mg ( d − y ) − mg (1 − h/d )(d − y )

= mg (h /d )(d − y ) and v = 2 gh(1 − y/d )

EVALUATE: This gives the correct results for y = 0 and for y = d . 7.62. IDENTIFY: Apply Eq. (7.14) to each stage of the motion. SET UP: Let y = 0 at the bottom of the slope. In part (a), Wother is the work done by friction. In part (b),

Wother is the work done by friction and the air resistance force. In part (c), Wother is the work done by the force exerted by the snowdrift. EXECUTE: (a) The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, K1 = mgh − W f = (60.0 kg)(9.8 N/kg)(65.0 m) − 10,500 J, or K1 = 38,200 J − 10,500 J = 27,720 J. Then v1 = 2 K1 2(27 ,720 J) = = 30.4 m/s. m 60 kg

(b) K 2 = K1 − (W f + Wair ) = 27,720 J − ( μk mgd + f air d ).

K 2 = 27,720 J − [(0.2)(588 N)(82 m) + (160 N)(82 m)] or K 2 = 27,720 J − 22,763 J = 4957 J. Then, v2 = 2K 2(4957 J) = = 12.9 m/s m 60 kg

(c) Use the Work-Energy Theorem to find the force. W = ΔK , F = K /d = (4957 J)/(2.5 m) = 2000 N. EVALUATE: In each case, Wother is negative and removes mechanical energy from the system. 7.63. IDENTIFY and SET UP: First apply ΣF = ma to the skier. Find the angle α where the normal force becomes zero, in terms of the speed v2 at this point. Then apply

the work-energy theorem to the motion of the skier to obtain another equation that relates v2 and α . Solve these two equations for α .

Let point 2 be where the skier loses contact with the snowball, as sketched in Figure 7.63a Loses contact implies n → 0. y1 = R, y2 = R cos α

Figure 7.63a

First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her acceleration is arad = v 2 /R, directed in towards the center of the snowball. EXECUTE: ΣFy = ma y

2 mg cos α − n = mv2 /R 2 But n = 0 so mg cos α = mv2 /R 2 v2 = Rg cos α

Figure 7.63b
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7-24

Chapter 7

Now use conservation of energy to get another equation relating v2 to α : K1 + U1 + Wother = K 2 + U 2

The only force that does work on the skier is gravity, so Wother = 0. 2 K1 = 0, K 2 = 1 mv2 2

U1 = mgy1 = mgR, U 2 = mgy2 = mgR cos α
2 Then mgR = 1 mv2 + mgR cos α 2 2 v2 = 2 gR(1 − cos α )

Combine this with the ΣFy = ma y equation:
Rg cos α = 2 gR (1 − cos α ) cos α = 2 − 2cos α 3cos α = 2 so cos α = 2/3 and α = 48.2° EVALUATE: She speeds up and her arad increases as she loses gravitational potential energy. She loses

7.64.

contact when she is going so fast that the radially inward component of her weight isn’t large enough to keep her in the circular path. Note that α where she loses contact does not depend on her mass or on the radius of the snowball. IDENTIFY: Initially the ball has all kinetic energy, but at its highest point it has kinetic energy and potential energy. Since it is thrown upward at an angle, its kinetic energy is not zero at its highest point. SET UP: Apply conservation of energy: K f + U f = Ki + U i . Let yi = 0, so yf = h, the maximum height. At this maximum height, vf , y = 0 and vf , x = vi, x , so vf = vi, x = (15 m/s)(cos60.0°) = 7.5 m/s. Substituting into conservation of energy equation gives EXECUTE: Solve for h: h =

1 mv 2 i 2 2

= mgh + 1 m(7.5 m/s)2 . 2

vi 2 − (7.5 m/s) (15 m/s)2 − (7.5 m/s) 2 = = 8.6 m. 2g 2(9.80 m/s 2 )

7.65.

EVALUATE: If the ball were thrown straight up, its maximum height would be 11.5 m, since all of its kinetic energy would be converted to potential energy. But in this case it reaches a lower height because it still retains some kinetic energy at its highest point. IDENTIFY and SET UP: yA = R

yB = yC = 0

Figure 7.65 (a) Apply conservation of energy to the motion from B to C: K B + U B + Wother = KC + U C . The motion is described in Figure 7.65. EXECUTE: The only force that does work on the package during this part of the motion is friction, so Wother = W f = f k (cos φ ) s = μ k mg (cos180°) s = -μ k mgs 2 K B = 1 mvB , KC = 0 2

U B = 0, U C = 0

Thus K B + W f = 0
1 mv 2 B 2

− μ k mgs = 0

μk =

μ2 B
2 gs

=

(4.80 m/s) 2 2(9.80 m/s 2 )(3.00 m)

= 0.392

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Potential Energy and Energy Conservation EVALUATE: The negative friction work takes away all the kinetic energy. (b) IDENTIFY and SET UP: Apply conservation of energy to the motion from A to B:

7-25

K A + U A + Wother = K B + U B
EXECUTE: Work is done by gravity and by friction, so Wother = W f . 2 K A = 0, K B = 1 mvB = 1 (0.200 kg)(4.80 m/s)2 = 2.304 J 2 2

U A = mgy A = mgR = (0.200 kg)(9.80 m/s 2 )(1.60 m) = 3.136 J, U B = 0

Thus U A + W f = K B
W f = K B − U A = 2.304 J − 3.136 J = -0.83 J
EVALUATE: W f is negative as expected; the friction force does negative work since it is directed 7.66.

opposite to the displacement. IDENTIFY: Apply Eq. (7.14) to the initial and final positions of the truck. SET UP: Let y = 0 at the lowest point of the path of the truck. Wother is the work done by friction. f r = μr n = μr mg cos β .

2 EXECUTE: Denote the distance the truck moves up the ramp by x. K1 = 1 mv0 , U1 = mgL sin α , K 2 = 0, 2

U 2 = mgx sin β and Wother = -μ r mgx cos β . From Wother = ( K 2 + U 2 ) − ( K1 + U1 ), and solving for x, x= K1 + mgL sin α (v 2 /2 g ) + L sin α = 0 . mg ( sin β + μr cos β ) sin β + μr cos β

EVALUATE: x increases when v0 increases and decreases when μr increases. 7.67. Fx = -α x − β x 2 , α = 60.0 N/m and β = 18.0 N/m 2 (a) IDENTIFY: Use Eq. (6.7) to calculate W and then use W = -ΔU to identify the potential energy function U ( x).

SET UP: WFx = U1 − U 2 = ∫
x

x2 F ( x) dx x1 x

Let x1 = 0 and U1 = 0. Let x2 be some arbitrary point x, so U 2 = U ( x ). EXECUTE: U ( x) = − ∫ Fx ( x ) dx = -∫ (−α x − β x 2 ) dx = ∫ (α x + β x 2 ) dx = 1 α x 2 + 1 β x3. 2 3 0 0 0 x x

EVALUATE: If β = 0, the spring does obey Hooke’s law, with k = α , and our result reduces to (b) IDENTIFY: Apply Eq. (7.15) to the motion of the object. SET UP: The system at points 1 and 2 is sketched in Figure 7.67.

1 kx 2 . 2

K1 + U1 + Wother = K 2 + U 2 The only force that does work on the object is the spring force, so Wother = 0.

Figure 7.67
2 EXECUTE: K1 = 0, K 2 = 1 mv2 2
2 3 U1 = U ( x1) = 1 α x1 + 1 β x1 = 1 (60.0 N/m)(1.00 m) 2 + 1 (18.0 N/m 2 )(1.00 m)3 = 36.0 J 2 3 2 3 2 3 U 2 = U ( x2 ) = 1 α x2 + 1 β x2 = 1 (60.0 N/m)(0.500 m)2 + 1 (18.0 N/m 2 )(0.500 m)3 = 8.25 J 2 3 2 3

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7-26

Chapter 7
2 Thus 36.0 J = 1 mv2 + 8.25 J 2

v2 =

2(36.0 J − 8.25 J) = 7.85 m/s 0.900 kg

7.68.

EVALUATE: The elastic potential energy stored in the spring decreases and the kinetic energy of the object increases. IDENTIFY: Mechanical energy is conserved on the hill, which gives us the speed of the sled at the top. After it leaves the cliff, we must use projectile motion. SET UP: Use conservation of energy to find the speed of the sled at the edge of the cliff. Let yi = 0 so

yf = h = 11.0 m. K f + U f = Ki + U i gives

1 mv 2 f 2

+ mgh = 1 mvi 2 or vf = vi 2 − 2 gh . Then analyze the 2

projectile motion of the sled: use the vertical component of motion to find the time t that the sled is in the air; then use the horizontal component of the motion with a x = 0 to find the horizontal displacement. EXECUTE: vf = (22.5 m/s) 2 − 2(9.80 m/s 2 )(11.0 m) = 17.1 m/s. yf = vi, yt + 1 a yt 2 gives 2

t=

2 yf 2(−11.0 m) = = 1.50 s. xf = vi, xt + 1 axt 2 gives xf = vi, xt = (17.1 m/s)(1.50 s) = 25.6 m. 2 ay -9.80 m/s 2

7.69.

EVALUATE: Conservation of energy can be used to find the speed of the sled at any point of the motion but does not specify how far the sled travels while it is in the air. IDENTIFY: Apply Eq. (7.14) to the motion of the block. SET UP: Let y = 0 at the floor. Let point 1 be the initial position of the block against the compressed

spring and let point 2 be just before the block strikes the floor. 2 EXECUTE: With U 2 = 0, K1 = 0, K 2 = U1. 1 mv2 = 1 kx 2 + mgh. Solving for v2 , 2 2 v2 = kx 2 (1900 N/m)(0.045 m)2 + 2 gh = + 2(9.80 m/s 2 )(1.20 m) = 7.01 m/s. m (0.150 kg)

7.70.

EVALUATE: The potential energy stored in the spring and the initial gravitational potential energy all go into the final kinetic energy of the block. IDENTIFY: Apply Eq. (7.14). U is the total elastic potential energy of the two springs. SET UP: Call the two points in the motion where Eq. (7.14) is applied A and B to avoid confusion with springs 1 and 2, that have force constants k1 and k2 . At any point in the motion the distance one spring is stretched equals the distance the other spring is compressed. Let + x be to the right. Let point A be the initial position of the block, where it is released from rest, so x1A = +0.150 m and x2 A = -0.150 m. EXECUTE: (a) With no friction, Wother = 0. K A = 0 and U A = K B + U B . The maximum speed is when

U B = 0 and this is at x1B = x2 B = 0, when both springs are at their natural length. 1 k x2 2 1 1A 2 2 2 2 + 1 k2 x2 A = 1 mvB . x1A = x2 A = (0.150 m) 2 , so 2 2

vB =

k1 + k2 2500 N/m + 2000 N/m (0.150 m) = (0.150 m) = 6.00 m/s. m 3.00 kg

(b) At maximum compression of spring 1, spring 2 has its maximum extension and vB = 0. Therefore, at

this point U A = U B . The distance spring 1 is compressed equals the distance spring 2 is stretched, and vice versa: x1A = - x2 A and x1B = - x2 B . Then U A = U B gives 1 (k + k ) x 2 2 1A 2 1 2 = 1 ( k1 + k2 ) x1B and 2

x1B = - x1A = -0.150 m. The maximum compression of spring 1 is 15.0 cm. EVALUATE: When friction is not present mechanical energy is conserved and is continually transformed between kinetic energy of the block and potential energy in the springs. If friction is present, its work removes mechanical energy from the system.

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Potential Energy and Energy Conservation 7.71.

7-27

IDENTIFY: Apply conservation of energy to relate x and h. Apply ΣF = ma to relate a and x. SET UP: The first condition, that the maximum height above the release point is h, is expressed as 1 kx 2 = mgh. The magnitude of the acceleration is largest when the spring is compressed to a distance x; at 2

this point the net upward force is kx − mg = ma, so the second condition is expressed as x = (m /k )( g + a ). EXECUTE: (a) Substituting the second expression into the first gives

1 ⎛m⎞ m( g + a ) 2 . k ⎜ ⎟ ( g + a) 2 = mgh, or k = 2 ⎝k⎠ 2 gh (b) Substituting this into the expression for x gives x = EVALUATE: When a → 0, our results become k =

2

2 gh . g+a

mg and x = 2h. The initial spring force is kx = mg 2h = mgh and sufficient potential energy is stored in the

and the net upward force approaches zero. But
7.72.

1 2 kx 2

spring to move the mass to height h. IDENTIFY: At equilibrium the upward spring force equals the weight mg of the object. Apply conservation of energy to the motion of the fish. SET UP: The distance that the mass descends equals the distance the spring is stretched. K1 = K 2 = 0, so

U1 (gravitational) = U 2 (spring) EXECUTE: Following the hint, the force constant k is found from mg = kd , or k = mg/d . When the fish falls from rest, its gravitational potential energy decreases by mgy; this becomes the potential energy of the 1 mg 2 y = mgy, or y = 2d . spring, which is 1 ky 2 = 1 (mg/d ) y 2 . Equating these, 2 2 2 d EVALUATE: At its lowest point the fish is not in equilibrium. The upward spring force at this point is ky = 2kd , and this is equal to twice the weight. At this point the net force is mg, upward, and the fish has 7.73.

an upward acceleration equal to g. IDENTIFY: Only conservative forces (gravity and the spring) act on the fish, so its mechanical energy is conserved. SET UP: Energy conservation tells us K1 + U1 + Wother = K 2 + U 2 , where Wother = 0. U g = mgy,

K = 1 mv 2 , and U spring = 1 ky 2 . 2 2
EXECUTE: (a) K1 + U1 + Wother = K 2 + U 2 . Let y be the distance the fish has descended, so y = 0.0500 m.

1 2 1 K1 = 0, Wother = 0, U1 = mgy, K 2 = mv2 , and U 2 = ky 2 . Solving for K2 gives 2 2 1 2 1 2 K 2 = U1 − U 2 = mgy − ky = (3.00 kg)(9.8 m/s )(0.0500 m) − (900 N/m)(0.0500 m)2 2 2

K 2 = 1.47 J − 1.125 J = 0.345 J. Solving for v2 gives v2 =

2K2 2(0.345 J) = = 0.480 m/s. 3.00 kg m

1 (b) The maximum speed is when K 2 is maximum, which is when dK 2 / dy = 0. Using K 2 = mgy − ky 2 2

gives

dK 2 mg (3.00 kg)(9.8 m/s 2 ) = mg − ky = 0. Solving for y gives y = = = 0.03267 m. At this y, 900 N/m dy k

1 K 2 = (3.00 kg)(9.8 m/s 2 )(0.03267 m) − (900 N/m)(0.03267 m) 2 . K 2 = 0.9604 J − 0.4803 J = 0.4801 J, 2 so v2 = 2K2 = 0.566 m/s. m EVALUATE: The speed in part (b) is greater than the speed in part (a), as it should be since it is the maximum speed.

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7-28 7.74.

Chapter 7 IDENTIFY: The spring obeys Hooke’s law. Gravity and the spring provide the vertical forces on the brick. The mechanical energy of the system is conserved. SET UP: Use K f + U f = Ki + U i . In part (a), setting yf = 0, we have yi = x, the amount the spring will

stretch. Also, since Ki = K f = 0,
EXECUTE: (a) x =

1 kx 2 2

= mgx. In part (b), yi = h + x, where h = 1.0 m.

2mg 2(3.0 kg)(9.80 m/s 2 ) = = 0.039 m = 3.9 cm. 1500 N/m k mg ⎛ 2hk (b) 1 kx 2 = mg (h + x), kx 2 − 2mgx − 2mgh = 0 and x = ⎜1 ± 1 + 2 k ⎜ mg ⎝ have x =

⎞ ⎟ . Since x must be positive, we ⎟ ⎠

7.75.

⎞ (3.0 kg)(9.80 m/s 2 ) ⎛ 2(1.0 m)(1500 N/m) ⎞ ⎜1 + 1 + ⎟ = 0.22 m = 22 cm ⎟= ⎟ ⎜ 1500 N/m 3.0 kg(9.80 m/s 2 ) ⎟ ⎠ ⎝ ⎠ EVALUATE: In part (b) there is additional initial energy (from gravity), so the spring is stretched more. (a) IDENTIFY and SET UP: Apply K A + U A + Wother = K B + U B to the motion from A to B. mg ⎛ 2hk ⎜1 + 1 + k ⎜ mg ⎝

2 EXECUTE: K A = 0, K B = 1 mvB 2

2 U A = 0, U B = U el ,B = 1 kxB , where xB = 0.25 m 2

Wother = WF = FxB
2 2 Thus FxB = 1 mvB + 1 kxB . (The work done by F goes partly to the potential energy of the stretched spring 2 2

and partly to the kinetic energy of the block.) 2 FxB = (20.0 N)(0.25 m) = 5.0 J and 1 kxB = 1 (40.0 N/m)(0.25 m) 2 = 1.25 J 2 2 2 Thus 5.0 J = 1 mvB + 1.25 J and vB = 2

2(3.75 J) = 3.87 m/s 0.500 kg

(b) IDENTIFY: Apply Eq. (7.15) to the motion of the block. Let point C be where the block is closest to the wall. When the block is at point C the spring is compressed an amount xC , so the block is

0.60 m − xC from the wall, and the distance between B and C is xB + xC . SET UP: The motion from A to B to C is described in Figure 7.75.

K B + U B + Wother = KC + U C
EXECUTE: Wother = 0
2 K B = 1 mvB = 5.0 J − 1.25 J = 3.75 J 2

(from part (a)) UB = = 1.25 J
1 2 kx 2 B

KC = 0 (instantaneously at rest at point
closest to wall)

UC =
Figure 7.75

1k 2

xC

2

Thus 3.75 J + 1.25 J = 1 k xC 2

2

2(5.0 J) = 0.50 m 40.0 N/m The distance of the block from the wall is 0.60 m − 0.50 m = 0.10 m. EVALUATE: The work (20.0 N)(0.25 m) = 5.0 J done by F puts 5.0 J of mechanical energy into the

xC =

system. No mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J.

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Potential Energy and Energy Conservation 7.76.

7-29

IDENTIFY: Apply Eq. (7.14) to the motion of the student. SET UP: Let x0 = 0.18 m, x1 = 0.71 m. The spring constants (assumed identical) are then known in terms

of the unknown weight w, 4kx0 = w. Let y = 0 at the initial position of the student. EXECUTE: (a) The speed of the brother at a given height h above the point of maximum compression is ⎛ x2 ⎞ ⎛ w⎞ (4k ) g 2 2 x1 − 2 gh = g ⎜ 1 − 2h ⎟ . Therefore, then found from 1 (4k ) x1 = 1 ⎜ ⎟ v 2 + mgh, or v 2 = 2 2 g ⎜x ⎟ w ⎝ ⎠ ⎝ 0 ⎠

v = (9.80 m/s 2 )((0.71 m) 2 /(0.18 m) − 2(0.90 m)) = 3.13 m/s, or 3.1 m/s to two figures. (b) Setting v = 0 and solving for h, h =
2 2kx1 x2 = 1 = 1.40 m, or 1.4 m to two figures. 2 x0 mg
2 ⎛ 0.53 m ⎞ x1 ( x0 + 0.53 m) 2 = = x0 ⎜1 + ⎟ will be x0 x0 x0 ⎠ ⎝ 2

(c) No; the distance x0 will be different, and the ratio

different. Note that on a planet with lower g, x0 will be smaller and h will be larger. EVALUATE: We are able to solve the problem without knowing either the mass of the student or the force constant of the spring. IDENTIFY: We can apply Newton’s second law to the block. The only forces acting on the block are gravity downward and the normal force from the track pointing toward the center of the circle. The mechanical energy of the block is conserved since only gravity does work on it. The normal force does no work since it is perpendicular to the displacement of the block. The target variable is the normal force at the top of the track. v2 SET UP: For circular motion ΣF = m . Energy conservation tells us that K A + U A + Wother = K B + U B , R

7.77.

where Wother = 0. U g = mgy and K = 1 mv 2 . 2
EXECUTE: Let point A be at the bottom of the path and point B be at the top of the path. At the bottom of v2 the path, n A − mg = m (from Newton’s second law). R

vA =

R 0.800 m ( n A − mg ) = (3.40 N − 0.49 N) = 6.82 m/s. Use energy conservation to find the speed 0.0500 kg m 1 mv 2 A 2 2 = 1 mvB + mg (2 R). Solving for vB gives 2

at point B. K A + U A + Wother = K B + U B , giving

vB = v 2 − 4 Rg = (6.82 m/s) 2 − 4(0.800 M)(9.8 m/s 2 ) = 3.89 m/s. Then at point B, Newton’s second law A gives nB + mg = m
2 vB . Solving for nB gives R ⎛ (3.89 m/s)2 ⎞ v2 nB = m B − mg = (0.0500 kg) ⎜ − 9.8 m/s 2 ⎟ = 0.456 N. ⎜ 0.800 m ⎟ R ⎝ ⎠ EVALUATE: The normal force at the top is considerably less than it is at the bottom for two reasons: the block is moving slower at the top and the downward force of gravity at the top aids the normal force in keeping the block moving in a circle. IDENTIFY: Applying Newton’s second law, we can use the known normal forces to find the speeds of the block at the top and bottom of the circle. We can then use energy conservation to find the work done by friction, which is the target variable. v2 SET UP: For circular motion ΣF = m . Energy conservation tells us that K A + U A + Wother = K B + U B , R

7.78.

where Wother is the work done by friction. U g = mgy and K = 1 mv 2 . 2

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7-30

Chapter 7 EXECUTE: Use the given values for the normal force to find the block’s speed at points A and B. At point A, v2 Newton’s second law gives n A − mg = m A . So R

vA =

R 0.500 m v2 ( n A − mg ) = (3.95 N − 0.392 N) = 6.669 m/s. Similarly at point B, nB + mg = m B . m R 0.0400 kg R 0.500 m ( nB + mg ) = (0.680 N + 0.392 N) = 3.660 m/s. Now apply the 0.0400 kg m

Solving for vB gives vB =

work-energy theorem to find the work done by friction. K A + U A + Wother = K B + U B .

Wother = K B + U B − K A .
1 1 Wother = (0.40 kg)(3.66 m/s) 2 + (0.04 kg)(9.8 m/s 2 )(1.0 m) − (0.04 kg)(6.669 m/s)2 . 2 2 Wother = 0.2679 J + 0.392 J − 0.8895 J = -0.230 J.

7.79.

EVALUATE: The work done by friction is negative, as it should be. This work is equal to the loss of mechanical energy between the top and bottom of the circle. IDENTIFY: U = mgh. Use h = 150 m for all the water that passes through the dam. SET UP: m = ρV and V = AΔh is the volume of water in a height Δh of water in the lake. EXECUTE: (a) Stored energy = mgh = ( ρ V ) gh = ρ A(1 m) gh.

stored energy = (1000 kg/m3 )(3.0 × 106 m 2 )(1 m)(9.8 m/s 2 )(150 m) = 4.4 × 1012 J. (b) 90% of the stored energy is converted to electrical energy, so (0.90)(mgh ) = 1000 kWh. (0.90) ρVgh = 1000 kWh. V = (1000 kWh)((3600 s)/(1 h)) (0.90)(1000 kg/m3 )(150 m)(9.8 m/s 2 ) = 2.7 × 103 m3.

Change in level of the lake: AΔh = Vwater . Δh =

7.80.

V 2.7 × 103 m3 = = 9.0 × 10−4 m. A 3.0 × 106 m 2 EVALUATE: Δh is much less than 150 m, so using h = 150 m for all the water that passed through the dam was a very good approximation. IDENTIFY and SET UP: The potential energy of a horizontal layer of thickness dy, area A, and height y is dU = (dm) gy. Let ρ be the density of water. EXECUTE: dm = ρ dV = ρ A dy, so dU = ρ Agy dy. The total potential energy U is

U = ∫ dU = ρ Ag ∫ y dy = 1 ρ Agh 2 . 2
0 0

h

h

A = 3.0 × 106 m 2 and h = 150 m, so U = 3.3 × 1014 J = 9.2 × 107 kWh EVALUATE: The volume is Ah and the mass of water is ρV = ρ Ah. The average depth is hav = h /2, so U = mghav . 7.81. IDENTIFY: Apply Eq. (7.15) to the motion of the block. SET UP: The motion from A to B is described in Figure 7.81.

Figure 7.81
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Potential Energy and Energy Conservation

7-31

The normal force is n = mg cosθ , so f k = μ k n = μ k mg cosθ .

y A = 0; yB = (6.00 m)sin 30.0° = 3.00 m K A + U A + Wother = K B + U B EXECUTE: Work is done by gravity, by the spring force, and by friction, so Wother = W f and

U = U el + U grav
2 K A = 0, K B = 1 mvB = 1 (1.50 kg)(7.00 m/s)2 = 36.75 J 2 2

U A = U el, A + U grav, A = U el, A , since U grav, A = 0 U B = U el, B + U grav, B = 0 + mgyB = (1.50 kg)(9.80 m/s 2 )(3.00 m) = 44.1 J

Wother = W f = ( f k cos φ ) s = μk mg cosθ (cos180°) s = -μk mg cosθ s Wother = -(0.50)(1.50 kg)(9.80 m/s 2 )(cos30.0°)(6.00 m) = -38.19 J

Thus U el, A − 38.19 J = 36.75 J + 44.10 J

U el, A = 38.19 J + 36.75 J + 44.10 J = 119 J
EVALUATE: U el must always be positive. Part of the energy initially stored in the spring was taken away

7.82.

by friction work; the rest went partly into kinetic energy and partly into an increase in gravitational potential energy. IDENTIFY: Only gravity does work, so apply Eq. (7.4). Use ΣF = ma to calculate the tension. SET UP: Let y = 0 at the bottom of the arc. Let point 1 be when the string makes a 45° angle with the vertical and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration arad = v 2 /r EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is mgl (1 − cos θ ), where l is the length of the string and θ is the angle the

string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so mgl (1 − cosθ ) = 1 mv 2 , or v = 2 gl (1 − cosθ ) = 2(9.80 m/s 2 )(0.80 m)(1 − cos 45°) = 2.1 m/s. 2 (b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to

the radial component of the weight, or mg cosθ = (0.12 kg)(9.80 m/s 2 ) cos 45° = 0.83 N. (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, 2 mg + mv2 /l = mg (1 + 2(1 − cos 45°)) = 1.9 N

7.83.

EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward. F = -αxy 2 ˆ, α = 2.50 N/m3 j IDENTIFY: F is not constant so use Eq. (6.14) to calculate W. F must be evaluated along the path. (a) SET UP: The path is sketched in Figure 7.83a.

ˆ dl = dxi + dyˆ j
F ⋅ dl = -α xy 2 dy

On the path, x = y so F ⋅ dl = -α y 3 dy
Figure 7.83a EXECUTE: W = ∫ F ⋅ dl = ∫
1 2
y 2 y 1 y ⎞ ⎛ 4 4 (−α y 3 ) dy = -(α /4) ⎜ y 4 ∫ 2 ⎟ = -(α /4)( y2 − y1 ) y ⎝ 1 ⎠

y1 = 0, y2 = 3.00 m, so W = - 1 (2.50 N/m3 )(3.00 m) 4 = -50.6 J 4

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7-32

Chapter 7 (b) SET UP: The path is sketched in Figure 7.83b.

Figure 7.83b

ˆ For the displacement from point 1 to point 2, dl = dxi , so F ⋅ dl = 0 and W = 0. (The force is perpendicular to the displacement at each point along the path, so W = 0.) j For the displacement from point 2 to point 3, dl = dyˆ, so F ⋅ dl = -α xy 2 dy. On this path, x = 3.00 m, so

F ⋅ dl = -(2.50 N/m3 )(3.00 m) y 2 dy = -(7.50 N/m 2 ) y 2 dy. EXECUTE: W = ∫ F ⋅ dl = -(7.50 N/m 2 ) ∫
2 3
y y

3

W = -(7.50 N/m 2 )

()
1 3

2

3 3 y 2 dy = -(7.50 N/m 2 ) 1 ( y3 − y2 ) 3

(3.00 m)3 = -67.5 J

7.84.

(c) EVALUATE: For these two paths between the same starting and ending points the work is different, so the force is nonconservative. IDENTIFY: Calculate the work W done by this force. If the force is conservative, the work is path independent. SET UP: W = ∫ P 2 P 1

F ⋅ dl .
P 2 P 1

EXECUTE: (a) W = ∫

Fy dy = C ∫

P 2 P 1

y 2 dy. W doesn't depend on x, so it is the same for all paths between

P1 and P2 . The force is conservative.

(b) W = ∫

P 2 P 1

Fx dx = C ∫

P 2 P 1

y 2 dx. W will be different for paths between points P1 and P2 for which y has

different values. For example, if y has the constant value y 0 along the path, then W = Cy 0 ( x2 − x1).

W depends on the value of y 0 . The force is not conservative. j EVALUATE: F = Cy 2 ˆ has the potential energy function U ( y ) = -

Cy 3 . We cannot find a potential 3

ˆ energy function for F = Cy 2i .
7.85. IDENTIFY: Use W = ∫
P 2 P 1

F ⋅ dl to calculate W for each segment of the path.

SET UP: F ⋅ dl = Fx dx = α xy dx EXECUTE: (a) The path is sketched in Figure 7.85. (b) (1): x = 0 along this leg, so F = 0 and W = 0. (2): Along this leg, y = 1.50 m, so

F ⋅ dl = (3.00 N/m) xdx, and W = (1.50 N/m)((1.50 m) 2 − 0) = 3.38 J (3) F ⋅ dl = 0, so W = 0 (4) y = 0, so F = 0 and W = 0. The work done in moving around the closed path is 3.38 J. (c) The work done in moving around a closed path is not zero, and the force is not conservative. EVALUATE: There is no potential energy function for this force.

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Potential Energy and Energy Conservation

7-33

Figure 7.85 7.86. IDENTIFY: Use Eq. (7.16) to relate Fx and U ( x). The equilibrium is stable where U ( x) is a local minimum and the equilibrium is unstable where U ( x) is a local maximum. SET UP: dU /dx is the slope of the graph of U versus x. K = E − U , so K is a maximum when U is a minimum. The maximum x is where E = U . EXECUTE: (a) The slope of the U vs. x curve is negative at point A, so Fx is positive (Eq. (7.16)). (b) The slope of the curve at point B is positive, so the force is negative. (c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m. (d) The curve at point C looks pretty close to flat, so the force is zero. (e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than U ( A), the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at any point where the potential energy is larger than U ( A). On the graph, that looks to be at

7.87.

about 2.2 m. (f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m. (g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C. EVALUATE: If E is less than U at point C, the particle is trapped in one or the other of the potential "wells" and cannot move from one allowed region of x to the other. IDENTIFY: K = E − U determines v( x ). SET UP: v is a maximum when U is a minimum and v is a minimum when U is a maximum. Fx = -dU /dx. The extreme values of x are where E = U ( x). EXECUTE: (a) Eliminating β in favor of α and x0 ( β = α /x0 ),

U ( x) =

α
x
2



β
x

=

2 x0

2 α x0

x

2



α
x0 x

=

⎢ ⎟ − ⎜ ⎟⎥ . 2 ⎜ x0 ⎢⎝ x ⎠ ⎝ x ⎠ ⎥ ⎣ ⎦

2 α ⎡⎛ x0 ⎞ ⎛ x0 ⎞ ⎤

⎛α ⎞ U ( x 0 ) = ⎜ 2 ⎟ (1 − 1) = 0. U ( x) is positive for x < x 0 and negative for x > x 0 ( α and β must be taken ⎜x ⎟ ⎝ 0⎠ as positive). The graph of U ( x) is sketched in Figure 7.87a. ⎟ . The proton moves in the positive x-direction, speeding up ⎟ ⎠ until it reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the square root in the expression for v ( x ) indicates that the particle will be found only in the region where U < 0, that is, x > x0 . The graph of v( x) is sketched in Figure 7.87b. (c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential 3 2 dU α ⎡ ⎛ x 0 ⎞ ⎛ x 0 ⎞ ⎤ dU = energy. This minimum occurs when = 0, or 3 ⎢ −2 ⎜ ⎟ + ⎜ ⎟ ⎥ = 0, dx x 0 ⎢ ⎜ x ⎟ ⎜ x ⎟ ⎥ dx ⎣ ⎝ ⎠ ⎝ ⎠ ⎦

⎛ 2α 2 (b) v( x) = - U = ⎜ 2 ⎜ mx m ⎝ 0

⎞ ⎛⎛ x ⎟ ⎜⎜ 0 ⎜ ⎟ ⎜⎝ x ⎠⎝

⎞ ⎛ x0 ⎟−⎜ ⎟ ⎜ x ⎠ ⎝

⎞ ⎟ ⎟ ⎠

2⎞

which has the solution x = 2 x 0 . U (2 x 0 ) = -

α
2 4 x0

, so v =

α
2 2mx 0

.

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7-34

Chapter 7

(d) The maximum speed occurs at a point where

dU = 0, and from Eq. (7.15), the force at this point dx

is zero.
(e) x1 = 3 x 0 , and U (3 x 0 ) = −


2 9 x0

.
⎞ α ⎛ ⎛ x ⎞2 x ⎞ ⎤ 2α ⎟ − ⎜ ⎜ 0 ⎟ − 0 ⎟⎥ = 2 ⎟ x2 ⎜ ⎜ x ⎟ x ⎟⎥ mx 0 ⎠ 0 ⎝⎝ ⎠ ⎠⎦ ⎛ ⎛ x ⎞ ⎛ x ⎞2 2 ⎞ ⎜ ⎜ 0 ⎟ − ⎜ 0 ⎟ − ⎟. ⎜⎝ x ⎠ ⎝ x ⎠ 9 ⎟ ⎝ ⎠

⎡ 2 2 ⎢ ⎛ -2 α ⎜ (U ( x1 ) − U ( x)) = v( x) = m m ⎢⎜ 9 x 2 0 ⎣⎝

The particle is confined to the region where U ( x) < U ( x1 ). The maximum speed still occurs at x = 2 x 0 , but now the particle will oscillate between x1 and some minimum value (see part (f)). (f) Note that U ( x) − U ( x1 ) can be written as

2 α ⎡⎛ x 0 ⎞ ⎛ x 0 ⎞ ⎛ 2 ⎞ ⎤ ⎢ ⎥

2 x0

α ⎡⎛ x0 ⎞ 1 ⎤ ⎡⎛ x0 ⎞ 2 ⎤ ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ = 2 ⎢⎜ ⎟ − ⎥ ⎢ ⎜ ⎟ − ⎥ , ⎜ x ⎟ ⎜ x ⎟ ⎝ 9 ⎠⎥ x ⎝ x ⎠ 3 ⎝ x ⎠ 3 ⎢⎝ ⎠ ⎝ ⎠ ⎦⎣ ⎦ 0 ⎣ ⎣ ⎦

which is zero (and hence the kinetic energy is zero) at x = 3 x 0 = x1 and x = 3 x 0 . Thus, when the particle 2 is released from x 0 , it goes on to infinity, and doesn’t reach any maximum distance. When released from x1 , it oscillates between 3 x 2 0

and 3x 0 .

EVALUATE: In each case the proton is released from rest and E = U ( xi ), where xi is the point where it

is released. When x i = x 0 the total energy is zero. When x i = x1 the total energy is negative. U ( x) → 0 as x → ∞, so for this case the proton can't reach x → ∞ and the maximum x it can have is limited.

Figure 7.87

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