Новое решение

Valter правка от 14 авг. 2026 г., 16:10

правка #20009 позже →

en/2.3.13.md +37 −0

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		+	### Statement
		+
		+	$2.3.13.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.3.13.$ A load of mass $m$ is slowly pulled up to a height $h$ along an inclined plane using a block and a cable. The work done during this process is $A$. Then the cable is released, and the load slides down. What velocity will it gain by the time it reaches the starting point?
		+
		+	![For problem $2.3.13$\|446x249, 40%](../../img/2.3.13/2.3.13.png)
		+
		+	### Solution
		+
		+	The difference between the work of the pulling force and the work of friction is equal to the change in potential energy:
		+	$$
		+	A - A_{\\text{f}} = mgh
		+	$$
		+	During the slide down from the top, friction performs the same amount of work. Let us write the law of conservation of energy:
		+	$$
		+	A_{\\text{f}} + \\frac{mv^2}{2} = mgh
		+	$$
		+	Substitute the work of friction to express the kinetic energy:
		+	$$
		+	\\frac{mv^2}{2} = 2mgh - A
		+	$$
		+	From this, we find the velocity the load will gain upon reaching the starting point:
		+	$$
		+	v = \\sqrt{4gh - 2A/m}
		+	$$
		+
		+	#### Answer
		+	$v = \\sqrt{4gh - 2A/m}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,37 @@
			### Statement

			$2.3.13.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.3.13.$ A load of mass $m$ is slowly pulled up to a height $h$ along an inclined plane using a block and a cable. The work done during this process is $A$. Then the cable is released, and the load slides down. What velocity will it gain by the time it reaches the starting point?

			![For problem $2.3.13$\|446x249, 40%](../../img/2.3.13/2.3.13.png)

			### Solution

			The difference between the work of the pulling force and the work of friction is equal to the change in potential energy:
			$$
			A - A_{\\text{f}} = mgh
			$$
			During the slide down from the top, friction performs the same amount of work. Let us write the law of conservation of energy:
			$$
			A_{\\text{f}} + \\frac{mv^2}{2} = mgh
			$$
			Substitute the work of friction to express the kinetic energy:
			$$
			\\frac{mv^2}{2} = 2mgh - A
			$$
			From this, we find the velocity the load will gain upon reaching the starting point:
			$$
			v = \\sqrt{4gh - 2A/m}
			$$

			#### Answer
			$v = \\sqrt{4gh - 2A/m}$

			#### Answer

			[Insert a concise answer or boxed result]