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)


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

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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

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

  • $\begingroup$ 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

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.


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