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Problem from an Indian entrance test

In this problem, if I use the formula for power as $\frac{F\cdot s}{t}$, then the answer is 37.5 W. But if I use the formula $F\cdot v$, then the answer is 75 W.

How is it possible that the man has two different powers? Which is the correct answer and why?

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closed as off-topic by Kyle Kanos, John Rennie, ZeroTheHero, Aaron Stevens, Jon Custer Feb 11 at 19:41

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    $\begingroup$ Can you show your work for the two methods? $\endgroup$ – The Photon Feb 11 at 15:58
  • $\begingroup$ Please use MathJax in this site so that others can lookup later even if the image gets deleted in the host website. $\endgroup$ – exp ikx Feb 11 at 16:03
  • $\begingroup$ $F\cdot v$ is true for a constant force moving an object at constant velocity, this is not true in your case. $\endgroup$ – Triatticus Feb 11 at 16:17
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    $\begingroup$ @Triatticus That is incorrect. $F\cdot v$ is valid for all cases, as it gives the instantaneous power being supplied. You don't need constant forces or constant velocities to use this equation $\endgroup$ – Aaron Stevens Feb 11 at 16:18
  • $\begingroup$ Instantaneous power is when you assume the changes are so small as to be approximately constant, it is only true when they are constant. It doesn't mean you can't use it but you'd need to apply calculus to do so $\endgroup$ – Triatticus Feb 11 at 16:21
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I wonder if you are not making a confusion bettween mean power and instantaneous power.

At the beginning, the speed is 0 and so the power is 0. And with time, the power increase. With your hypothesis, $v=Ft/m$ and so $P={{F}^{2}}t/m$

The work is $W(t)=\int\limits_{0}^{t}{Pdt}=P=\frac{1}{2}{{F}^{2}}t/m$ and the mean power $\left\langle P \right\rangle =\frac{W(t)}{t}=\int\limits_{0}^{t}{Pdt}=\frac{1}{2}{{F}^{2}}t/m=\frac{1}{2}P(t)$ That is why you find a factor $1/2$

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  • $\begingroup$ I didn't understand what you did there but I got the idea. My concepts are not clear on mean and instantaneous powers. $\endgroup$ – Virendra Sankpal Feb 11 at 16:50
  • $\begingroup$ Maybe the difference betweem mean velocity and instantaneous would be more familiar to you. The velocity of the boat increase linearly with time. It will be maximal at the end. To compute the mean velocity, you have to compute the distance (integral of the speed) and divide by the total time. Clearly, the mean velocity will be less that the maximal velocity. $\endgroup$ – Vincent Fraticelli Feb 12 at 7:24
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P=w/t=(F*s)/t this is average power w=F•s diffrentiating both side wrt time

dw/dt=d(F•s)/dt . when force is constant

dw/dt=F•ds/dt=F•v. now dw/dt is rate of doing work then by defination it is power

p=F•v . this gives instantenous poweras ds/dt gives velocity of that instant. your question asks for average power.so 37.5watts is correct

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  • $\begingroup$ I think the question asks for instantaneous power. Read it again. $\endgroup$ – Virendra Sankpal Feb 11 at 16:51
  • $\begingroup$ no because it says at end of third second from beginning of motiom $\endgroup$ – user221619 Feb 11 at 16:53
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P=F v is for constant velocity v. ur velocity is accelerating at a constant rate, so u must convert the velocity back to an average v. The way to do that is divide the sum of initial velocity and final velocity by 2. If u draw out a v t Graph it would be clearer as to why divide by 2. anyway, Vinitial is 0ms-1 and Vfinal is a x 3s = .75ms-1 so avg velocity = (0+0.75)/2=0.375ms-1 now we apply P=F v = (100)(0.375)=37.5W, which agrees with ur other result

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