Question

A bullet of mass 0.04 kg moving with speed of 90 M per second enters a heavy wooden block and it stopped after a distance of 60 cm what is the average resistive force exerted by the block on the bullet

Answer 1

use formula
F of bullet= M × (v^2 - u^2)÷2 S i.e distance
F= 0.04 × (0- 8100)÷(2×0.6)
F= 2700 N
force exerted by bullet = force exerted by block
Force by block = 2700 N