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. [tex]\Delta[/tex]Uinternal = Friction Force* Distance =(12.7106N)(.180)(.230m) = 0.52621884J From here, i do not know what to do. Here was what I attempted: KE=[tex]\Delta[/tex]Uinternal 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.