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+### Statement
+
+$2.3.5.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.3.5.$ 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:
+$$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$
+
+#### Answer
+$$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]