$2.3.5.$ The force acting on a projectile of mass $m$ in a gun barrel increases uniformly from zero to $F_0$ over a barrel length $l_1$, remains constant over a barrel length $l_2$, and finally decreases uniformly to zero over a barrel length $l_3$. What is the velocity of the projectile upon exiting the barrel?
Solution
From the law of conservation of mechanical energy, the kinetic energy of the projectile equals the total work done by the force: $$E_{\text{mec}} = A_1 + A_2 + A_3 = \frac{mv^2}{2} \quad \text{(a)}$$
The work parts $A_1, A_2$ and $A_3$ can be found as the respective areas under the force-displacement graph: $$A_1 = \frac{1}{2}F_0 \cdot l_1; \quad A_2 = F_0 \cdot l_2; \quad A_3 = \frac{1}{2}F_0 \cdot l_3$$
Substitute these into equation (a): $$E_{\text{mec}} = \frac{mv^2}{2} = F_0 \cdot \frac{l_1 + 2l_2 + l_3}{2}$$
Express the velocity $v$ of the projectile upon exiting the barrel: $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$
