Physics Chapter 7 solutions

# Thus the tension is t 230 kg980 ms 2 cos 195 2391 n

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Thus, the tension is T = (230 kg)(9.80 m/s 2 )/cos 19.5° = 2391 N and F = (2391 N) sin 19.5° = 797 N. An alternative approach based on drawing a vector triangle (of forces) in the final situation provides a quick solution. (b) Since there is no change in kinetic energy, the net work on it is zero. (c) The work done by gravity is W F d mgh g g ´ G G , where h = L (1 – cos T ) is the vertical component of the displacement. With L = 12.0 m, we obtain W g = –1547 J, which should be rounded to three significant figures: –1.55 kJ. (d) The tension vector is everywhere perpendicular to the direction of motion, so its work is zero (since cos 90° = 0). (e) The implication of the previous three parts is that the work due to G F is – W g (so the net work turns out to be zero). Thus, W F = – W g = 1.55 kJ. (f) Since G F does not have constant magnitude, we cannot expect Eq. 7-8 to apply.

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CHAPTER 7 294 58. (a) The force of the worker on the crate is constant, so the work it does is given by W F d Fd F G G cos I , where G F is the force, G d is the displacement of the crate, and I is the angle between the force and the displacement. Here F = 210 N, d = 3.0 m, and I = 20°. Thus, W F = (210 N) (3.0 m) cos 20° = 590 J. (b) The force of gravity is downward, perpendicular to the displacement of the crate. The angle between this force and the displacement is 90° and cos 90° = 0, so the work done by the force of gravity is zero. (c) The normal force of the floor on the crate is also perpendicular to the displacement, so the work done by this force is also zero. (d) These are the only forces acting on the crate, so the total work done on it is 590 J. 59. (a) We set up the ratio 50 1 1 3 km 1 km megaton F H G I K J E / and find E = 50 3 | 1 u 10 5 megatons of TNT. (b) We note that 15 kilotons is equivalent to 0.015 megatons. Dividing the result from part (a) by 0.013 yields about ten million bombs. 60. (a) In the work-kinetic energy theorem, we include both the work due to an applied force W a and work done by gravity W g in order to find the latter quantity. 30 J (100 N)(1.8 m)cos 180 a g g K W W W ' ± leading to 2 2.1 10 J g W u . (b) The value of W g obtained in part (a) still applies since the weight and the path of the child remain the same, so 2 2.1 10 J g W . ' u . 61. One approach is to assume a “path” from G r i to G r f and do the line-integral accordingly. Another approach is to simply use Eq. 7-36, which we demonstrate: 4 3 2 3 (2 ) (3) f f i i x y x y x y W F dx F dy x dx dy ´ ´ ± ± ³ ³ ³ ³ with SI units understood. Thus, we obtain W = 12 J – 18 J = – 6 J.
295 62. (a) The compression of the spring is d = 0.12 m. The work done by the force of gravity (acting on the block) is, by Eq. 7-12, W mgd 1 0 25 0 29 ( . . kg) 9.8 m / s (0.12 m) J. 2 c h (b) The work done by the spring is, by Eq. 7-26, W kd 2 2 1 2 1 2 250 18 ´ ´ ´ N / m) (0.12 m) J. 2 ( . (c) The speed v i of the block just before it hits the spring is found from the work-kinetic energy theorem (Eq. 7-15): ' K mv W W i ´ ± 0 1 2 2 1 2 which yields 1 2 ( 2)( ) ( 2)(0.29 J 1.8 J) 3.5 m/s.

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