Question

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?

Answer 1

Answer:

Explanation:

Conservation of momentum during the collision

0.023(230) + 2.0(0.0) = 0.023(170) + 2.0v

v = 0.69 m/s

The initial block kinetic energy will be converted to friction work

½mv² = Fd = μmgd

½(2.0)(0.69²) = 0.15(2.0)(9.8)d

d = 0.1619387... m

d = 16 cm