Work & Engery Problem?

Question

A bullet having a mass of 5g is fired horizontally into a wooden block of 2.5 kg mass that is sitting at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is μk = 0.25. The bullet comes to rest in the wooden block, which moves a distance of 2 m along the surface. 1) How much work did the bullet do in pushing the block? 2) What was the speed of the bullet when it hit the block? 3) How much kinetic energy was absorbed by the block (i.e. how much energy was not available to push the block)?

Answer 1

1) the work done by friction is
m*g*μk*d
here
(.005+2.5)*9.81*0.25*2
12.3 J
This is the work done by the bullet

2)since momentum is conserved,
first find the speed after collision
.5*2.505*v^2=12.3
=3.13 m/s

then
.005*v=2.505*3.13
v=1568 m/s

kinetic energy of the bullet was
.5*.005*1568^2

6146 J

3) The block absorbed 6132 J

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