A 3.6 kg block of wood rests on a long tabletop. A 28.2 g bullet moving horizont

Question

A 3.6 kg block of wood rests on a long tabletop. A 28.2 g bullet moving horizontally with a speed of 150 m/s is shot into the block and sticks in it. The block then slides 39.5 cm along the table and stops. Find the magnitude of the frictional force between the block and the table.

Explanation / Answer

Initial momentum = final momentum =>(3.6*0)+((28.2/1000)*150) = (3.6+(28.2/1000))*U =>U = 1.165867372 m/s......This is initial velocity Final velocity = 0 m/s as it comes to rest. S = (39.5/100) m V^2 = U^2 + 2aS =>a = (1.165867372^2)/(2*(39.5/100)) = 1.72056548 m/s^2 Frictional force = ma = (3.6+(28.2/1000))*1.72056548 = 6.242555675 N

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