A 79.7-g arrow is fired horizontally from rest. The bowstring exerts an average

Question

A 79.7-g arrow is fired horizontally from rest. The bowstring exerts an average force of 83.9 N on the arrow over a distance of 0.815 m. With what speed does the arrow leave the bow? Relative to the ground, what is the gravitational potential energy of a 49.9-kg person who is at the top of the Sears Tower, a height of 443 m above the ground? When a 0.0607-kg golf ball takes off after being hit, its speed is 32.7 m/s. (a) How much work is done on the ball by the club? (b) Assume that the force of the golf club acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of 0.0441 m. Ignore the weight of the ball and determine the average force applied to the ball by the club. A 2.26-kg rock is released from rest at a height of 28.0 m. Ignore air resistance and determine (a) the kinetic energy at 28.0 m, (b) the gravitational potential energy at 28.0 m, (c) the total mechanical energy at 28.0 m, (d) the kinetic energy at 0 m, (e) the gravitational potential energy at 0 m, and (f) the total mechanical energy at 0 m.

Explanation / Answer

2) F = ma

83.9 = 0.0797 * a

a = 1052.69 m/s2

v^2 - u^2 = 2aS

v^2 - 0 = 2*1052.69*0.815

v = 41.42 m/s

3) PE = mgh

= 49.9*9.8*443 = 216635.86 J

4) a) work done = change in KE

= 0.5*mv^2 = 0.5*0.0607*32.7^2 = 32.45 J

b) W = F.S

F = W/S = 32.45 / 0.0441 = 735.9 N

5) a) KE at 28 m = 0 because the rock is at rest

b) PE = mgh = 2.26*9.8*28 = 620.144 J

c) Total mechanical energy = KE + PE = 0 + 620.144 J

d) KE at 0 = PE at 28 = 620.144 J

e) PE at 0 = mgh = 0

f) total energy remains same(by conservation of energy principle) = 620.144 J

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