A bullet and a block (Momentum)

[Chandasouk]

A bullet of mass 7.00 g is fired horizontally into a wooden block of mass 1.29 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.180. The bullet remains embedded in the block, which is observed to slide a distance 0.230 m along the surface before stopping.

What was the initial speed of the bullet?



Okay, first off, I know the block is sliding after the collision so that is internal energy.

$$\Delta$$U_internal = Friction Force* Distance

=(12.7106N)(.180)(.230m) = 0.52621884J

From here, i do not know what to do.

Here was what I attempted:

KE=$$\Delta$$U_internal

so

.5(1.297kg)v^2 = 0.52621884J

Velocity of the block was 0.9007m/s

Then I used Conservation of Momentum

Pi = Pf

.007kgV = (1.297kg)(.9007m/s)

V=166.9m/s


I don't know, seems slow for a bullet.

 

[AEM]

Yes, it seems slow for a bullet, but then 23 cm doesn't seem very far for the block to slide. Your approach looks correct to me.