A bullet of mass 40 g moving with a speed of 90 ms-1 enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is

Here, u = 90 ms-1,  v = 0     m = 40 g   4001000kg = 0.04 kgs = 60 cm = 0.6 mUsing v2-u2 = 2as(0)2- (90)2 = 2a ×0.6a = (90)22×0.6 = -6750 ms-2-ve sign shows the retardation.The average resistive force exerted by block on the bullet is F = m ×a = (0.04 kg) (6750 ms-2) = 270 N