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<title>A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?</title>
$2.3.41.$ A part of the mass $m$ is detached from the load hanging on the spring of stiffness $k$. To what height will the remaining part of the cargo then rise?
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" width="100" />
<figcaption>
For problem $2.3.41$
</figcaption>
</figure>
</center>
<p>
</p>
<h3>Solution</h3>
<p>
Since there is no interaction with external factors other than breakaway, all the work goes into changing the length of the spring. $E_k=0$!
</p><p>
Small change in the pontential energy of gravity
$$
dA = mg \, dx
$$
Small change in the pontential energy of the spring
$$
dE = kx \, dx
$$
Law of conservation of energy
$$
dA = dE
$$
$$
mg \, dx = kx \, dx
$$
$$
mg \int dx = k \int x \, dx
$$
From where
$$
mg \, x = \frac{kx^2}{2}\Rightarrow\boxed{x = \frac{2mg}{k}}
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
<meta name="description" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
<meta property="og:title" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
<meta property="og:title" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
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<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og:description" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
<meta property="og:description" content="A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?">
<title>A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?</title>
<title>A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?</title>
$2.3.41.$ A part of the mass $m$ is detached from the load hanging on the spring of stiffness $k$. To what height will the remaining part of the cargo then rise?
$2.3.41.$ A part of the mass $m$ is detached from the load hanging on the spring of stiffness $k$. To what height will the remaining part of the cargo then rise?
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" width="100" />
loading="lazy" width="100" />
<figcaption>
<figcaption>
For problem $2.3.41$
For problem $2.3.41$
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Since there is no interaction with external factors other than breakaway, all the work goes into changing the length of the spring. $E_k=0$!
Since there is no interaction with external factors other than breakaway, all the work goes into changing the length of the spring. $E_k=0$!
</p><p>
</p><p>
Small change in the pontential energy of gravity
Small change in the pontential energy of gravity
$$
$$
dA = mg \, dx
dA = mg \, dx
$$
$$
Small change in the pontential energy of the spring
Small change in the pontential energy of the spring
$$
$$
dE = kx \, dx
dE = kx \, dx
$$
$$
Law of conservation of energy
Law of conservation of energy
$$
$$
dA = dE
dA = dE
$$
$$
$$
$$
mg \, dx = kx \, dx
mg \, dx = kx \, dx
$$
$$
$$
$$
mg \int dx = k \int x \, dx
mg \int dx = k \int x \, dx
$$
$$
From where
From where
$$
$$
mg \, x = \frac{kx^2}{2}\Rightarrow\boxed{x = \frac{2mg}{k}}
mg \, x = \frac{kx^2}{2}\Rightarrow\boxed{x = \frac{2mg}{k}}
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
