Новое решение

Valter правка от 30 авг. 2026 г., 19:52

правка #20450 позже →

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		+	### Statement
		+
		+	$2.3.17.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.3.17.$ A cart of mass $m$ rolls with velocity $v$ along rails forming a horizontal circular path of radius $R$. A worker runs after it and begins to stop it by pulling a cable tied to the cart with a force $F$ at an angle of $\pi - \alpha$ to the direction of the cart's velocity. How many revolutions around the circle will the cart make before stopping? Neglect friction.
		+
		+	### Solution
		+
		+	The cart will stop when all its initial kinetic energy is spent on the work done by the stopping force $F$.
		+	The projection of the force acting directly against the velocity of the cart is $F \cos\alpha$ (since the angle with the velocity vector is $\pi - \alpha$).
		+
		+	Equating the initial kinetic energy to the magnitude of the work done:
		+	$$\frac{mv^2}{2} = \int F \cos\alpha \, ds$$
		+	Since the force is constant, we can take it out of the integral and integrate over the entire path distance $S = 2\pi n R$, where $n$ is the number of full revolutions:
		+	$$\frac{mv^2}{2} = F \cos\alpha \cdot 2\pi n R$$
		+
		+	From this, we express the number of revolutions $n$ around the circle before stopping:
		+	$$n = \frac{mv^2}{4\pi F R \cos\alpha}$$
		+	where $n$ is the required number of full revolutions of the cart until it stops.
		+
		+	#### Answer
		+	$$n = \frac{mv^2}{4\pi F R \cos\alpha}$$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.3.17.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.3.17.$ A cart of mass $m$ rolls with velocity $v$ along rails forming a horizontal circular path of radius $R$. A worker runs after it and begins to stop it by pulling a cable tied to the cart with a force $F$ at an angle of $\pi - \alpha$ to the direction of the cart's velocity. How many revolutions around the circle will the cart make before stopping? Neglect friction.

			### Solution

			The cart will stop when all its initial kinetic energy is spent on the work done by the stopping force $F$.
			The projection of the force acting directly against the velocity of the cart is $F \cos\alpha$ (since the angle with the velocity vector is $\pi - \alpha$).

			Equating the initial kinetic energy to the magnitude of the work done:
			$$\frac{mv^2}{2} = \int F \cos\alpha \, ds$$
			Since the force is constant, we can take it out of the integral and integrate over the entire path distance $S = 2\pi n R$, where $n$ is the number of full revolutions:
			$$\frac{mv^2}{2} = F \cos\alpha \cdot 2\pi n R$$

			From this, we express the number of revolutions $n$ around the circle before stopping:
			$$n = \frac{mv^2}{4\pi F R \cos\alpha}$$
			where $n$ is the required number of full revolutions of the cart until it stops.

			#### Answer
			$$n = \frac{mv^2}{4\pi F R \cos\alpha}$$

			#### Answer

			[Insert a concise answer or boxed result]