A 2 kg block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of 150 ms-1 is shot into the block and sticks to it. The block then slides 2.7 m along the table top and comes to a stop. The force of friction between the block and the table is

Velocity of block just after collision,                 v=5×10-3×1502+5×10-3                                  (from conservation of linear momentum)                    = 0.374 ms-1Let F be the force of friction, then work done against friction = initial kinetic energyor         F×2.7=12×2.005×(0.374)2⇒F=0.052 N