Edits to “Statement”, “Solution”, “Answer”
en/2.3.7.md
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| ### Statement | |||
| − | $2.3.7.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
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| − | ### Statement | ||
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| $2.3.7.$ A window blind of mass $m = 1\text{ kg}$ and length $L = 2\text{ m}$ is rolled up into a thin roller above the window. What is the minimum work spent on this? Neglect friction. | |||
| ### Solution | |||
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| + | |||
| To roll the blind into a thin roller at the top, we need to raise its center of mass. Initially, for a uniform rectangular blind of length $L$, the center of mass is located precisely in the middle, at a distance of $\frac{L}{2}$ from the top. | |||
| Therefore, the minimum required work is equal to the change in the potential energy of the blind's center of mass: | |||
| $$A = mg\frac{L}{2}$$ | |||
| Substituting the given values: | |||
| $$A = 1 \cdot 10 \cdot \frac{2}{2} = 10\text{ J}$$ | |||
| @@ -20,7 +16,3 @@Solution | |||
| #### Answer | |||
| $$A = 10\text{ J}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,15 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $2.3.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.7.$ A window blind of mass $m = 1\text{ kg}$ and length $L = 2\text{ m}$ is rolled up into a thin roller above the window. What is the minimum work spent on this? Neglect friction. | $2.3.7.$ A window blind of mass $m = 1\text{ kg}$ and length $L = 2\text{ m}$ is rolled up into a thin roller above the window. What is the minimum work spent on this? Neglect friction. | ||
| ### Solution | ### Solution | ||
|  | |||
| To roll the blind into a thin roller at the top, we need to raise its center of mass. Initially, for a uniform rectangular blind of length $L$, the center of mass is located precisely in the middle, at a distance of $\frac{L}{2}$ from the top. | To roll the blind into a thin roller at the top, we need to raise its center of mass. Initially, for a uniform rectangular blind of length $L$, the center of mass is located precisely in the middle, at a distance of $\frac{L}{2}$ from the top. | ||
| Therefore, the minimum required work is equal to the change in the potential energy of the blind's center of mass: | Therefore, the minimum required work is equal to the change in the potential energy of the blind's center of mass: | ||
| $$A = mg\frac{L}{2}$$ | $$A = mg\frac{L}{2}$$ | ||
| Substituting the given values: | Substituting the given values: | ||
| $$A = 1 \cdot 10 \cdot \frac{2}{2} = 10\text{ J}$$ | $$A = 1 \cdot 10 \cdot \frac{2}{2} = 10\text{ J}$$ | ||
| @@ -20,7 +16,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$A = 10\text{ J}$$ | $$A = 10\text{ J}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||