Answer to Question #77745 in Mechanics | Relativity for Muhammad Fiaz

Question

A bullet of mass m = 1 g is fired with speed v = 300 m/s into a wooden block(fixed to the floor). The bullet penetrates into the block for a distance s = 20 cm then is stops and remains embedded in the block. Calculate the frictional force that the wood exert on the bullet, supposed to be constant.
A: P = 450 N
B: P = 300 N
C: P = 150 N
D: P = 225 N
E: P = 90 N

Answer 1

Let’s apply the Work-Energy theorem. According to it, the change in the kinetic energy of the bullet is equal to the work done by the frictional force to stop the bullet:
〖KE〗_f-〖KE〗_i=W_fr,
1/2 mv_f^2-1/2 mv_i^2=-F_fr s,
-1/2 mv_i^2=-F_fr s,
F_fr=(mv_i^2)/2s=(0.001 kg∙(300 〖ms〗^(-1) )^2)/(2∙0.20 m)=225 N.