Added 2.4.20 3.3.8

astrosander правка от 30 авг. 2024 г., 18:46

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en/2.4.20.md +124 −0

		@@ -0,0 +1,124 @@
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		+	<body style="">
		+	<header style="text-align:center;">
		+	<h2>Solutions of Savchenko Physics Textbook</h2>
		+	<p class="author">
		+	Aliaksandr Melnichenka <br/>
		+	October 2023
		+	</p>
		+	</header>
		+
		+	<h3 id="back-link"><a href="../#2.4">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$2.4.20^*.$ A dog of mass m is tied with a leash of length $L$ to a sledge of mass $M > m$. At the initial moment, the dog is next to the sled. What is the greatest distance a dog can move the sledge in one jerk, if the coefficients of friction of the dog's paws and the sled runners on the horizontal surface are the same?
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="statement.png"
		+	loading="lazy" width="180" />
		+	<figcaption>
		+	For problem $2.4.20^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	The dog pulls the sled, performing work equal to
		+	\[
		+	A = \int_0^L F \, ds
		+	\]
		+	Where:
		+	\[
		+	F = \mu mg
		+	\]
		+	So we get:
		+	\[
		+	A = \mu mg L
		+	\]
		+	According to the work-energy theorem, we know that:
		+	\[
		+	A = \Delta K = \frac{mv^2}{2}
		+	\]
		+	After the dog has run a distance $L$, the leash tethered to the sled will stretch. This scenario can be analyzed as an inelastic collision.
		+	\[
		+	mv = (M + m)u
		+	\]
		+	In this scenario, the work done against the frictional force represents the energy lost by the system, which results in the deceleration of the sled-dog system. This loss of kinetic energy can be quantified using the work-energy principle:
		+	\[
		+	\frac{(M + m) u^2}{2} = \mu (M - m) g x
		+	\]
		+	From there we can derive:
		+	\[
		+	x = \frac{(M + m) u^2}{2 \mu g (M - m)}
		+	\]
		+
		+	We already know, that:
		+	\[
		+	u = \frac{m}{M + m}v
		+	\]
		+
		+	Finally:
		+	\[
		+	\boxed{x = \frac{m^2}{M^2 - m^2}L}
		+	\]
		+
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$x = \frac{m^2}{M^2 - m^2}L$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	physicshub<br>
		+	</p>
		+
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		+	<br>
		+	<p>
		+	<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
		+	</p>
		+	<p>
		+	<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
		+	</p>
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			<meta property="og:title" content="Savchenko Solutions">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="A website with solutions to physics problems from Savchenko Textbook">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>Savchenko Solutions</title>
			<link rel="stylesheet" href="https://savchenko-physics.github.io/css/css-latex/style.css">
			<link rel="icon" href="https://savchenko-physics.github.io/img/logo.png" type="image/png">
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			<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
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			ignoreClass: 'html'
			},
			showProcessingMessages: false,
			messageStyle: 'none'
			});
			</script>
			</head>
			<body style="">
			<header style="text-align:center;">
			<h2>Solutions of Savchenko Physics Textbook</h2>
			<p class="author">
			Aliaksandr Melnichenka <br/>
			October 2023
			</p>
			</header>

			<h3 id="back-link"><a href="../#2.4">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$2.4.20^*.$ A dog of mass m is tied with a leash of length $L$ to a sledge of mass $M > m$. At the initial moment, the dog is next to the sled. What is the greatest distance a dog can move the sledge in one jerk, if the coefficients of friction of the dog's paws and the sled runners on the horizontal surface are the same?
			</p>
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" width="180" />
			<figcaption>
			For problem $2.4.20^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			The dog pulls the sled, performing work equal to
			\[
			A = \int_0^L F \, ds
			\]
			Where:
			\[
			F = \mu mg
			\]
			So we get:
			\[
			A = \mu mg L
			\]
			According to the work-energy theorem, we know that:
			\[
			A = \Delta K = \frac{mv^2}{2}
			\]
			After the dog has run a distance $L$, the leash tethered to the sled will stretch. This scenario can be analyzed as an inelastic collision.
			\[
			mv = (M + m)u
			\]
			In this scenario, the work done against the frictional force represents the energy lost by the system, which results in the deceleration of the sled-dog system. This loss of kinetic energy can be quantified using the work-energy principle:
			\[
			\frac{(M + m) u^2}{2} = \mu (M - m) g x
			\]
			From there we can derive:
			\[
			x = \frac{(M + m) u^2}{2 \mu g (M - m)}
			\]

			We already know, that:
			\[
			u = \frac{m}{M + m}v
			\]

			Finally:
			\[
			\boxed{x = \frac{m^2}{M^2 - m^2}L}
			\]

			</p>

			<h4>Answer</h4>
			<p>
			$$x = \frac{m^2}{M^2 - m^2}L$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			physicshub<br>
			</p>


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			</p>
			<p>
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