A 25 g bullet strikes and becomes embedded in a 1.55 kg block of wood placed on a horizontal surface...?

Question

A 25 g bullet strikes and becomes embedded in a 1.55 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 0.45, and the impact drives the block a distance of 9.5 m before it comes to rest, what was the muzzle speed of the bullet?

__________m/s

Answer 1

Work done W on the block is equal to kinetic energy Ke of the bullet.

W=fs=u m2g s where m2= m(block) + m(bullet)
Ke=0.5 m1 V^2
0.5 m1 V^2=u m2g s
V=sqrt(2 u g s (m2/m1))
V= sqrt(2 x 0.45 x 9.81 x 9.5m (1.55+0.025)/.025)
V= 24 m/s