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.
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.
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.
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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)
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...