Правка разделов «Statement», «Solution», «Answer»
en/2.3.5.md
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| ### Statement | |||
| − | $2.3.5.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $2.3.5.$ 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? | |||
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| ### Solution | |||
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| 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: | |||
| $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | |||
| @@ -24,7 +22,3 @@Solution | |||
| #### Answer | |||
| $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### 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? | $2.3.5.$ 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? | ||
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| ### Solution | ### Solution | ||
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| From the law of conservation of mechanical energy, the kinetic energy of the projectile equals the total work done by the force: | From the law of conservation of mechanical energy, the kinetic energy of the projectile equals the total work done by the force: | ||
| $$E_{\text{mec}} = A_1 + A_2 + A_3 = \frac{mv^2}{2} \quad \text{(a)}$$ | $$E_{\text{mec}} = A_1 + A_2 + A_3 = \frac{mv^2}{2} \quad \text{(a)}$$ | ||
| The work parts $A_1, A_2$ and $A_3$ can be found as the respective areas under the force-displacement graph: | The work parts $A_1, A_2$ and $A_3$ can be found as the respective areas under the force-displacement graph: | ||
| $$A_1 = \frac{1}{2}F_0 \cdot l_1; \quad A_2 = F_0 \cdot l_2; \quad A_3 = \frac{1}{2}F_0 \cdot l_3$$ | $$A_1 = \frac{1}{2}F_0 \cdot l_1; \quad A_2 = F_0 \cdot l_2; \quad A_3 = \frac{1}{2}F_0 \cdot l_3$$ | ||
| Substitute these into equation (a): | Substitute these into equation (a): | ||
| $$E_{\text{mec}} = \frac{mv^2}{2} = F_0 \cdot \frac{l_1 + 2l_2 + l_3}{2}$$ | $$E_{\text{mec}} = \frac{mv^2}{2} = F_0 \cdot \frac{l_1 + 2l_2 + l_3}{2}$$ | ||
| Express the velocity $v$ of the projectile upon exiting the barrel: | Express the velocity $v$ of the projectile upon exiting the barrel: | ||
| $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | ||
| @@ -24,7 +22,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | $$v = \sqrt{\frac{F_0(l_1 + 2l_2 + l_3)}{m}}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||