-1
$\begingroup$

Suppose a block is attached to a compressed spring, it is released so moves under spring force of magnitude $kx$, let it moves from $-a$ (negative since displacement is negative for a compressed string) to $0$ (equilibrium position), then work done by spring is given by integral (upper limit: $0$ and lower limit: $-a$) so

$$W = (1/2)\times k \times 0 - (1/2) \times k \times a^2$$

This is a negative value, however the block gains velocity. Am I missing something or the integration limits are invalid? Please explain through the concept of work-energy theorem (not potential energy)

$\endgroup$

closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, sammy gerbil, Kyle Kanos, glS Aug 11 '18 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Emilio Pisanty, glS
If this question can be reworded to fit the rules in the help center, please edit the question.

3
$\begingroup$

You have signs confused. You can see this by realising that, if the spring is uncompressed at $x=0$, then it will be pulling or pushing towards $x=0$. So the force it exerts is $-kx$, not $kx$: if it exerted $kx$ then $x=0$ is an unstable equilibrium position, and any perturbation from it would result in the spring exploding violently (consider the case when $x$ is large: there is a force of $kx$ trying to make $x$ larger.

I think the thing I've described above is an important trick: I am terrible at sign errors, but like most physicists I have reasonably good intuition. So the trick is to think about what happens physically. In particular, if the force were $kx$, what happens to the spring physically if $x$ is large? Once you do that you immediately see that the force can't be $kx$ because of the whole exploding problem: it has to be $-kx$.


Here is the working in detail.

So, the initial displacement is $-a$, the final displacement is $0$. The force is $-kx$ where $x$ is displacement. So the integral you need to do is

$$\begin{align} \int\limits_{x=-a}^0 -kx\,dx &= \left. \frac{-kx^2}{2}\right\vert_{x=-a}^0\\ &= 0 - \left(\frac{-k(-a)^2}{2}\right)\\ &= 0 + \frac{k(-a)^2}{2}\\ &= \frac{ka^2}{2} \end{align}$$

$\endgroup$
  • $\begingroup$ You forgot a 1/2 factor :) Man this problem is really getting people today haha $\endgroup$ – Aaron Stevens Aug 6 '18 at 15:43
  • $\begingroup$ My answer is also just wrong: I'm going to nuke it. $\endgroup$ – tfb Aug 6 '18 at 15:44
  • $\begingroup$ @AaronStevens: I think it's right now. $\endgroup$ – tfb Aug 6 '18 at 15:51
  • $\begingroup$ Yes thank you! I was in the middle of typing up my own answer. I am surprised at how tripped up people are getting here haha $\endgroup$ – Aaron Stevens Aug 6 '18 at 15:52
  • 2
    $\begingroup$ @James: I've amended my answer (again!) with a physical intuition paragraph, which tries to explain why you can see, physically, why the force can't be what the questioner thought: the system is horrible unstable (and unstable in a way which involves infinite amounts of energy!) $\endgroup$ – tfb Aug 6 '18 at 16:00
0
$\begingroup$

Consider a spring with one end fixed and the other end which can move in the $\hat x$ direction.
When the spring is at its natural length the free end of the spring is at position $0\hat x$.
If the spring is extended the $x >0$ and if the spring is compressed then $x<0$.

The force that the free end of the spring exerts on a body attached to it is $\vec F = - k \vec x =- k x \hat x$ where $k$ is the spring constant.

If $x>0$ then the force that the extended spring exerts on the block is in the $-\hat x$ direction and if $x<0$ thn the force that the compressed spring exerts on the block is in the $\hat i$ direction.

Suppose that the end of the spring starts at position $x \hat x$ and finishes at position $0\hat i$ then the work done by the spring is

$$\int_x^0 \vec F \cdot d\vec x = \int_x^0 (-kx\hat x) \cdot (dx \hat x)= \int_x^0 -kx\, dx = \frac 12 kx^2$$

Now I have done it this way without making any assumption as to whether the spring is extended or compressed ie whether $x$ is positive or negative.
You will note that if $x=-a$ the work done by the spring is positive because $(-a)^2$ is positive and in your example the elastic potential energy of the spring, $\dfrac12 k a^2$, is converted into an increase in the kinetic energy of the block by that amount.

If the block started from rest it would stop again when $x=a$, the elastic potential energy of the spring again being $\dfrac12 k a^2$, and then the block would retrace its steps undergoing simple harmonic motion in the process.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.