A 15.0 gram bullet strikes and becomes embedded in a 1.24 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.280, and the impact drives the block a distance of 11.0 m before it comes to rest, what was the muzzle speed of the bullet?

Answer 1

Vb = vel of bullet
initial momentum = .015 Vb
=final momentum = 1.255 Vi
friction force = 1.255(.280)(9.81)

initial KE = (1/2)(1.255) Vi^2
work done = inital Ke = 1.255(.280)(9.81)(11)
solve that for Vi
then
Vb = (1,255/.015)Vi