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Spring in a railway carriage

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A man stretches a spring attached to the front wall of a railway carriage over a distance $l$ in a uniformly moving train. During this time the train covered the distance $L$. What work will be performed by the man in a coordinate system related to the Earth? When the man stretches the spring he moves in a direction opposite to that of the train.

I got $\frac{1}{2}k(L−l)^2$ but answer given is $\frac{1}{2}kl(l−L)$

My attempt:

The elongation after any time t is $X=x−vt$ where $x$ is the length pulled by the man. So $dX=dx−vdt$. Now, change in potential energy(kXdX) is the work done by man (I feel that this step is wrong). So I got the work done by as the above answer. I feel this method is wrong because the railway carriage also does some work, and therefore change in potential energy might equal the work done by the man and the carriage.

asked Nov 6, 2016 in Physics Problems by Mathman (120 points)
edited Nov 6, 2016 by Einstein
The answer given in your book is incorrect. I've attached the original problem along with its solution below.

2 Answers

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Best answer

With due respect, I think @Einstein's answer is incorrect as he simply found the work done by the man on the spring and not the "work done by the man". In that case we need to find both work done by man on spring and on moving train.

I would like to point out some facts.

When viewed from ground frame :

1) Work done by the man on the spring is decreased from $\frac{kl^2}{2}$ to $\frac{k(l-L)(l)}{2}$.

2) Work done by man on train increases from $0$ to $\frac{k(L)(l)}{2}$.

3) Work done by train on spring increases from $0$ to $\frac{k(L)(l)}{2}$.

4) However, total work done on the spring by both man and train is $\frac{kl^2}{2}$.

5) Also, total work done by man on spring and on train is $\frac{kl^2}{2}$.

The problem is from a famous book (whose pdf I'm linking here). See problem number 139.

I'm attaching the solution here.

answered Nov 6, 2016 by xxxx (160 points)
edited Nov 9, 2016 by Einstein
0 votes

Let's start with the equation for work,

$W = \int_{s_0}^{s_1} F.dx$

For a spring, $F = kx$ where $x$ is the distance the spring is displaced.

In the Earth's frame, let the initial position of the end of the spring (part being held) be 0. The final position is thus $-(L-l) = l-L$.

Therefore we need to evaluate the integral,

$W = \int_{0}^{l-L} F.dx$

We know that when the end of the spring is at position $l-L$ (final position), the spring has been displaced by a net amount of $l$ (as the net displacement of the spring is the same in both frames. This is because both the front and back of the spring are moving together in the train so the final "stretch" of the spring is the same whether viewing from Earth or from within the train). Therefore using proportionality, we know that when the end of the spring is at position $x$ the spring is displaced by an amount $\frac{l}{l-L}\times x$.

The force being applied when the position of the end of the spring is $x$ is thus given by $F = k\times displacement = k\times \frac{l}{l-L}\times x = \frac{klx}{l-L}$

We then substitue this back into the equation for work and integrate from $x=0$ to $x=l-L$,

$W = \int_{0}^{l-L} \frac{klx}{l-L}dx = \frac{kl}{l-L}\int_{0}^{l-L} xdx = \frac{1}{2}kl(l-L)$

answered Nov 6, 2016 by Einstein (1,496 points)
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