Question

A 20 g bullet moving horizontally with a speed v hits a 2-kg block of wood resting on a table. After hitting the block the bullet is embedded in the block of wood and the block and the bullet moves a distance of 40 cm before it stops. If the coefficient of friction between the block and the table is 0.2 find the initial speed v of the block.

Answer 1

mass of block with bullet = 2+.020 =2.02 kg
weight of block with bullet = 2.02*9.81 = 19.8 N
friction force = 19.8*.2 = 3.96 N
work done by friction = 3.96 * .40 = 1.59 J
that is kinetic energy of block with bullet
1.59 = (1/2)(2.02) v^2
v = 1.25 m/s for block with bullet
now conservation of momentum
2.02 * 1.25 = .020 * v
v = 127 m/s