Question

A 4.0-g bullet leaves the muzzle of a rifle with a speed of 340 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.75-m-long barrel of the rifle?

Answer 1

Given:

$m=0.004\:\rm kg$

$v_0=0,\quad v=340\:\rm m/s$

$d=0.75\:\rm m$

The energy-work theorem says

$$\frac{mv^2}{2}-\frac{mv_0^2}{2}=Fd$$

Thus, the force exerted onj the bullet

$$F=\frac{mv^2}{2d}=\frac{0.004*340^2}{2*0.75}=310\:\rm N$$