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## Homework Statement

A 12g bullet is fired in a 87.8g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 9.8m before coming to rest. The acceleration of gravity is 9.8 m/s^2

If the coefficient of friction between the surface and the block is .5, find the speed of the bullet before impact.

Answer: 81.5033 m/s

## Homework Equations

Normal Force = mg

Work = Distance * Force

Another one with velocity in the variable

## The Attempt at a Solution

The farthest I got was finding the frictional force that opposes the motion of the block.

Normal Force = (.012+.0878)(9.8) = .97804 in the upward direction

Friction Force = (.5)(.97804) = .48902

I suppose I could turn that into work, which would be:

Work = (.48902)(9.8) = 4.79240 in the westward direction

That gives me the force that opposes the motion of the block, but I don't think that gets me anywhere closer to the velocity. This is where I am stuck, and although I know the answer, I will have to know how to do this for the test. Any guidance will be much appreciated.