A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.

a. use work and energy to find the magnitude of the force that stops the bullet.

b. assuming that the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment the bullet stops moving.

A) Work energy relation;

Work = ΔKE; work done = Force * distance, while, Kinetic energy = 1/2 MV²

F. s = 1/2mv²

F * 4*10^-2 = 1/2 * 5 * 10^-3 * (600) ²

F = 900/0.04

= 22500 N

Therefore, force is 22500 N

b) From newton's second law of motion;

F = Ma

Thus; a = F/m

= 22500 / (5*10^-3)

= 4,500,000 m/s²

But v = u-at

0 = 600 - 4500,000 t

t = 1.33 * 10^-4 seconds