Правка разделов «Statement», «Solution», «Answer»
en/2.3.15.md
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| ### Statement | |||
| − | $2.3.15.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $2.3.15.$ What minimum work must be done to raise a long uniform pole of length $l$ and mass $m$ lying on the ground to a vertical position? | |||
| ### Solution | |||
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| For a uniform pole, its center of mass is located exactly in the middle. When lying on the ground, the height of its center of mass is $0$. When placed vertically, the center of mass is raised to a height of $h = \frac{l}{2}$. | |||
| The minimum work done is equal to the increase in the potential energy of the center of mass: | |||
| $$A = mgh = mg\frac{l}{2}$$ | |||
| #### Answer | |||
| $$A = mg\frac{l}{2}$$ | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### Statement | ||
| $2.3.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.15.$ What minimum work must be done to raise a long uniform pole of length $l$ and mass $m$ lying on the ground to a vertical position? | $2.3.15.$ What minimum work must be done to raise a long uniform pole of length $l$ and mass $m$ lying on the ground to a vertical position? | ||
| ### Solution | ### Solution | ||
|  | |||
| For a uniform pole, its center of mass is located exactly in the middle. When lying on the ground, the height of its center of mass is $0$. When placed vertically, the center of mass is raised to a height of $h = \frac{l}{2}$. | For a uniform pole, its center of mass is located exactly in the middle. When lying on the ground, the height of its center of mass is $0$. When placed vertically, the center of mass is raised to a height of $h = \frac{l}{2}$. | ||
| The minimum work done is equal to the increase in the potential energy of the center of mass: | The minimum work done is equal to the increase in the potential energy of the center of mass: | ||
| $$A = mgh = mg\frac{l}{2}$$ | $$A = mgh = mg\frac{l}{2}$$ | ||
| #### Answer | #### Answer | ||
| $$A = mg\frac{l}{2}$$ | $$A = mg\frac{l}{2}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||