Новое решение

Valter правка от 30 авг. 2026 г., 19:18

правка #20432 позже →

en/2.3.7.md +26 −0

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		+	### 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.
		+
		+	### 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.
		+
		+	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:
		+	$$A = 1 \cdot 10 \cdot \frac{2}{2} = 10\text{ J}$$
		+
		+	#### Answer
		+	$$A = 10\text{ J}$$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### 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.

			### 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.

			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:
			$$A = 1 \cdot 10 \cdot \frac{2}{2} = 10\text{ J}$$

			#### Answer
			$$A = 10\text{ J}$$

			#### Answer

			[Insert a concise answer or boxed result]