A 4 g bullet leaves the muzzle of a riffle with a speed of 350 m s-1. What force is exerted on the bullet when it is travelling down the 0.85 m long barrel of the rifle?

Answer 1

F = m*a

V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = ((350)^2-0)/1.7 = 72,059 m/s^2

F = 0.004 * 72059 = 288 N.