a bullet of mass 0.04kg moving with a speed of 90m/s enter a heavy wooden block is stopped after a distance of 60cm.What is the avg resistive force exertes by the block or the bullet

1 answer

a=(V^2-Vo^2)/2d=(0-90^2)/1.2m=-6750m/s^2

F = m*a = 0.04 * -6750 = -270 N.