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Question: A block of mass $m$ is lying on a smooth horizontal surface when a force $F$ is applied on it horizontally. If it's kinetic energy as a function of displacement $s$ is given by $k s^2$, then find the acceleration.

(a) $\frac{ks}{2m}$ $~~~$ (b) $\frac{ks}{m}$ $~~~$ (c) $\frac{ks}{4m}$ $~~~$ (d) $\frac{2 k s}{m}$

Attempt: At first it may seem a very simple physics question but two different approaches make the analysis complex. The approaches are as follows.

By the Work-energy theorem, work done equals the gain in kinetic energy, that is, $mas = ks^2$, so that $a = \frac{ks}{m}$.

Another approach is as follows, $\frac{1}{2} mv^2 = ks^2$ so that $v^2 = 2 \frac{k s^2}{m}$ and differentiating both sides w.r.t $s$ results in $v \cdot \frac{dv}{ds} = a = \frac{2 ks}{m}$.

Why do I get different answers when both the paths seem to be correct?

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  • $\begingroup$ Warm Welcome to Physics stack exchange @Apoorv Mishra. Your question is very difficult to read due to be written in plain text, I suggest you learn how to write in mathjax so you can more easily communicate with other members of the site in mathematics line here $\endgroup$ – Buraian Mar 4 at 8:12
  • $\begingroup$ Be careful with your uppercase and lowercase symbols. It's better to keep it consistent. Also, be careful with spaces in derivations to clearly show which parts belong together and which don't. Also, what is "M. a. S" and "w.r.t S. v."? Finally, how does the differentiation in the last step imply "dv/dS= a"? Usually, acceleration is the time derivative of speed. Would you mind outlining this differentiation a bit more - I suspect this is where something might be off. $\endgroup$ – Steeven Mar 4 at 8:13
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    $\begingroup$ In your first attempt, you can not write work done = F.S instead to calculate work you need to integrate F.ds over the required range, because F is not constant in the given situation, which we can identify from the fact that acceleration is dependent upon the displacement of the particle. $\endgroup$ – Chaitanya Garg Mar 4 at 8:33
  • $\begingroup$ I apologize for my writing error since I'm new on physics stack exchange and not quite acquainted with the common writing style here. I will learn soon. $\endgroup$ – Apoorv Mishra Mar 4 at 8:48
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Your first approach is wrong. You assume that the work is $F\cdot S$, which is only valid iff $F$ is a constant force. However, the force that satisfies your scenario is not constant.

The easiest way to show this would be take the negative gradient of the potential energy which gives the force. If you haven't learned this, then you can try solving for the equation of motion:

You have that the $\mathrm{KE} = kS^2$ therefore,$$\dfrac12mv^2=kS^2\implies\sqrt{\dfrac m2}v=\sqrt k S\\ \implies\sqrt{\dfrac m2}\dfrac{dS}{dt}=\sqrt k S$$

If you solve for $S(t)$, you will realize it's an exponential function. Its second time derivative is certainly not constant, so the $W=Fd$ relationship won't hold.

Your second approach should work fine because you do not assume the force is constant.

$$\dfrac12 mv^2 = kS^2 \implies \dfrac{d}{dt}\left( \dfrac12 mv^2 \right)= \dfrac{d}{dt} \left( kS^2 \right)\implies mv\dfrac{dv}{dt} = 2kS\dfrac{dS}{dt}$$

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  • $\begingroup$ No, I've written v. dv/dS = a $\endgroup$ – Apoorv Mishra Mar 4 at 8:53
  • $\begingroup$ Thank you for your answer I've understood the concept behind it I just missed out the fact that acceleration ain't constant here. $\endgroup$ – Apoorv Mishra Mar 4 at 8:54
  • $\begingroup$ yes, sorry, I misread your solution. By the chain rule, $$a=\dfrac{dv}{dt} = \dfrac{dv}{ds}\dfrac{ds}{dt} = \dfrac{dv}{ds}v$$ so you are right to say that $a=v\dfrac{dv}{ds}$ $\endgroup$ – user256872 Mar 4 at 8:55

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