Answer
The average net force exerted on the bullet is $4.3kN$
Work Step by Step
1) Find the acceleration $a$ of the bullet
We have the bullet's initial speed $v_0=0$, its final speed $v=715m/s$ and the time $\Delta t=2.5\times10^{-3}s$. Thus, $$v=v_0+at$$ $$a=\frac{v-v_0}{t}=2.86\times10^{5}m/s^2$$
2) Use Newton's 2nd Law to find the net force exerted on the bullet
We have $m_{bullet}=15g=1.5\times10^{-2}kg$
$$\sum F=m_{bullet}\times a=4.3\times10^3N=4.3kN$$
