A 4.2 g bullet leaves the muzzle of a rifle with

Question

a speed of 372.4 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.5 m length of
the barrel of the rifle?
Answer in units of N

Answer 1

V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = (372.4^2-0)/1 = 138,682 m/s^2

F = m * a = 0.0042 * 138,682 = 582.5 N.