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en/2.3.3.md
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| + | ### Statement | ||
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| + | $2.3.3.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | ### Statement | ||
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| + | $2.3.3.$ To test equipment under conditions of extreme acceleration and weightlessness, a container with it is tossed to a height of $125\text{ m}$ by a pneumatic piston device located at the bottom of a vacuum shaft. With what force does the piston act while tossing the container, if it extends by a length $h = 1\text{ m}$ in the process, and the mass of the container with the equipment is $m = 2\text{ t}$? | ||
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| + | ### Solution | ||
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| + | The piston device imparts a certain amount of kinetic energy to the equipment, which, according to the law of conservation of mechanical energy (since the shaft is a vacuum), converts into potential energy. Therefore, the work done by the force of the piston device is equal to the change in potential energy: | ||
| + | $$Fh = mgH \implies F = mg\frac{H}{h}$$ | ||
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| + | Let's substitute the numerical values: | ||
| + | $$F = 2 \cdot 10^3\text{ kg} \cdot 10\text{ }\frac{\text{N}}{\text{kg}} \cdot \frac{125\text{ m}}{1\text{ m}} = 2.5 \cdot 10^6\text{ N}$$ | ||
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| + | #### Answer | ||
| + | $F = 2.5 \cdot 10^6\text{ N}$ | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.3.3.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.3.$ To test equipment under conditions of extreme acceleration and weightlessness, a container with it is tossed to a height of $125\text{ m}$ by a pneumatic piston device located at the bottom of a vacuum shaft. With what force does the piston act while tossing the container, if it extends by a length $h = 1\text{ m}$ in the process, and the mass of the container with the equipment is $m = 2\text{ t}$? | |||
| ### Solution | |||
| The piston device imparts a certain amount of kinetic energy to the equipment, which, according to the law of conservation of mechanical energy (since the shaft is a vacuum), converts into potential energy. Therefore, the work done by the force of the piston device is equal to the change in potential energy: | |||
| $$Fh = mgH \implies F = mg\frac{H}{h}$$ | |||
| Let's substitute the numerical values: | |||
| $$F = 2 \cdot 10^3\text{ kg} \cdot 10\text{ }\frac{\text{N}}{\text{kg}} \cdot \frac{125\text{ m}}{1\text{ m}} = 2.5 \cdot 10^6\text{ N}$$ | |||
| #### Answer | |||
| $F = 2.5 \cdot 10^6\text{ N}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||