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en/2.3.18.md
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| + | ### Statement | ||
| + | |||
| + | $2.3.18.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $2.3.18^*.$ A rope is tied to a sled and passed over a crossbar of a gate of height $h$. A boy sitting on the sled starts pulling the rope, tensioning it with a force $T$. What speed will he acquire passing under the crossbar? The initial length of the tensioned part of the rope is $2l$, and the mass of the boy with the sled is $m$. Neglect friction. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Let the horizontal distance from the gate to the sled be $x$, and the length of the rope segment from the crossbar to the sled be $L$. The horizontal speed of the sled is $v = - \frac{dx}{dt}$ (the negative sign indicates that the coordinate $x$ decreases as the sled moves forward). | ||
| + | |||
| + | Since the rope is both tied to the sled and pulled by the boy sitting on it, the total external force acting on the system along the rope is $2T$. | ||
| + | The horizontal projection of this force drives the sled forward: | ||
| + | $$F_x = 2T \cos\alpha = 2T \frac{x}{L}$$ | ||
| + | |||
| + | According to Newton's Second Law for the horizontal axis: | ||
| + | $$m \frac{dv}{dt} = 2T \frac{x}{L}$$ | ||
| + | |||
| + | Instead of calculating the work done by a variable force, we can find the relation between $\frac{x}{L}$ and the speed using the kinematic constraint. From the Pythagorean theorem: | ||
| + | $$L^2 = x^2 + h^2$$ | ||
| + | |||
| + | Differentiating both sides with respect to time $t$ (keeping in mind that $h = \text{const}$): | ||
| + | $$2L \frac{dL}{dt} = 2x \frac{dx}{dt}$$ | ||
| + | |||
| + | Substitute the horizontal speed $\frac{dx}{dt} = -v$: | ||
| + | $$L \frac{dL}{dt} = -x v \implies \frac{x}{L} = -\frac{1}{v} \frac{dL}{dt}$$ | ||
| + | |||
| + | Now, substitute this relation back into the equation of motion: | ||
| + | $$m \frac{dv}{dt} = 2T \left( -\frac{1}{v} \frac{dL}{dt} \right)$$ | ||
| + | $$m v \frac{dv}{dt} = -2T \frac{dL}{dt}$$ | ||
| + | $$m v \, dv = -2T \, dL$$ | ||
| + | |||
| + | Integrate both sides from the initial state (speed $v = 0$, segment length $L = l$) to the final state directly under the crossbar (speed $v = V$, segment length $L = h$): | ||
| + | $$\int_{0}^{V} m v \, dv = \int_{l}^{h} -2T \, dL$$ | ||
| + | $$\frac{mV^2}{2} = -2T(h - l) = 2T(l - h)$$ | ||
| + | |||
| + | Solving for the final speed $V$: | ||
| + | $$V = \sqrt{\frac{4T(l - h)}{m}}$$ | ||
| + | |||
| + | #### Answer | ||
| + | $$V = \sqrt{\frac{4T(l - h)}{m}}$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.3.18.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.18^*.$ A rope is tied to a sled and passed over a crossbar of a gate of height $h$. A boy sitting on the sled starts pulling the rope, tensioning it with a force $T$. What speed will he acquire passing under the crossbar? The initial length of the tensioned part of the rope is $2l$, and the mass of the boy with the sled is $m$. Neglect friction. | |||
| ### Solution | |||
| Let the horizontal distance from the gate to the sled be $x$, and the length of the rope segment from the crossbar to the sled be $L$. The horizontal speed of the sled is $v = - \frac{dx}{dt}$ (the negative sign indicates that the coordinate $x$ decreases as the sled moves forward). | |||
| Since the rope is both tied to the sled and pulled by the boy sitting on it, the total external force acting on the system along the rope is $2T$. | |||
| The horizontal projection of this force drives the sled forward: | |||
| $$F_x = 2T \cos\alpha = 2T \frac{x}{L}$$ | |||
| According to Newton's Second Law for the horizontal axis: | |||
| $$m \frac{dv}{dt} = 2T \frac{x}{L}$$ | |||
| Instead of calculating the work done by a variable force, we can find the relation between $\frac{x}{L}$ and the speed using the kinematic constraint. From the Pythagorean theorem: | |||
| $$L^2 = x^2 + h^2$$ | |||
| Differentiating both sides with respect to time $t$ (keeping in mind that $h = \text{const}$): | |||
| $$2L \frac{dL}{dt} = 2x \frac{dx}{dt}$$ | |||
| Substitute the horizontal speed $\frac{dx}{dt} = -v$: | |||
| $$L \frac{dL}{dt} = -x v \implies \frac{x}{L} = -\frac{1}{v} \frac{dL}{dt}$$ | |||
| Now, substitute this relation back into the equation of motion: | |||
| $$m \frac{dv}{dt} = 2T \left( -\frac{1}{v} \frac{dL}{dt} \right)$$ | |||
| $$m v \frac{dv}{dt} = -2T \frac{dL}{dt}$$ | |||
| $$m v \, dv = -2T \, dL$$ | |||
| Integrate both sides from the initial state (speed $v = 0$, segment length $L = l$) to the final state directly under the crossbar (speed $v = V$, segment length $L = h$): | |||
| $$\int_{0}^{V} m v \, dv = \int_{l}^{h} -2T \, dL$$ | |||
| $$\frac{mV^2}{2} = -2T(h - l) = 2T(l - h)$$ | |||
| Solving for the final speed $V$: | |||
| $$V = \sqrt{\frac{4T(l - h)}{m}}$$ | |||
| #### Answer | |||
| $$V = \sqrt{\frac{4T(l - h)}{m}}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||