Physics help please
posted by Iamdumb on .
A 13.2 gm bullet is fired into a 12.0 kg wood block that is at rest on a wood table. The block, with the embedded bullet, slides 5.00 cm across the table. Assuming that the local acceleration due to gravity is 9.80m/s2 and that the coefficient of kinetic friction between the block and the table is 0.190, what was the speed of the bullet (in m/s)?

Use the distance that the bullet and block travel after impact (X = 0.0500 m)to get the initial kinetic energy of bullet and block together.
(Friction force)*(distance)
= (1/2)(M+m) Vo^2
M + m = 12.0132 kg
Friction force = (0.190)(M+m) g
= 22.37 N
Solve for Vo
Once you have Vo, compute the initial momentum (M+m) Vo
That equals the initial momentum of the bullet. Divide that by bullet mass m for the bullet's velocity.