A 0.001 kg bullet is fired from a gun and lodges inside a wooden block of mass 0.2 kg. The block and bullet then slide on a rough floor with a coefficient of kinetic friction μk = 0.4 before coming to rest after sliding a distance of 3 m.

The initial velocity of the bullet was?

Question

A 0.001 kg bullet is fired from a gun and lodges inside a wooden block of mass 0.2 kg. The block and bullet then slide on a rough floor with a coefficient of kinetic friction μk = 0.4 before coming to rest after sliding a distance of 3 m.

The initial velocity of the bullet was?

anonymous · Answer

simply apply this formula :
workdone by the friction force=change in kinetic energy
meu*normal force*d=1/2mu^2-0
meu*(M+m)*g*d=1/2mu^2
u=sqrt((2*meu*(M+m)*g*d)/m)...where M=0.2kg,m=0.001kg,d=3m,meu=0.4,g=9.8m/sec^2...find u which is initial velocity of bullet..