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

© 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.

7-1

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...