A 4.3 g bullet leaves the muzzle of a rifle with a speed of 306 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.89 m long barrel of the rifle?

Answer 1

a = (Vf^2 - Vo^2) / 2d,
a = ((306)^2 - 0) / 1.78 = 52,605m/s^2.

F = ma = 0.0043kg * 52,605 = 226N.