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A car of 1200 kg acquires a velocity of $ 25$ m/s from rest in 8 seconds. What amount of power should be given to car by the engine?

Procedure-1

Work done, \begin{align} W= \frac{1}{2}mv^2- \frac{1}{2}mu^2 \end{align} \begin{align} W=\frac{1}{2}×1200×(25^2)-0\end{align} Power,\begin{align} P=\frac{W}{t} =\frac{375000}{8}=46875\end{align}

Procedure-2

Power,\begin{align} P=\frac{dW}{dt}=\frac{d}{dt}{\int Fvdt}=Fv=mav=1200×\frac{25}{8}×25=93750\end{align}

Why is procedure-2 wrong?

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  • $\begingroup$ Is there, in fact, such a thing as an amount of power? As opposed to, say, a peak power level... $\endgroup$
    – DJohnM
    Jun 5 at 22:51

3 Answers 3

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The second derivation gives the wrong answer because it is trying to compute a different quantity. The first method is calculating the average power, while the second method is calculating the power at time $t$. I'll derive the second method more carefully in the hope that it will clarify what is happening.

It should look like: \begin{align} \frac{d}{dt}W(t)&=\frac{d}{dt}\int_0^tdt'F(t')v(t')\\ &=m\frac{d}{dt}\int_0^tdt'\frac{dv}{dt'}v(t') \end{align} Note that I took great care to make the dummy index $t'$ in the integral different from the "regular" variable $t$. Now, can we solve this integral without knowing what $v(t)$ looks like? Surprisingly, we can. By doing so we essentially derive method 1. Start by noticing that $\frac{d}{dt}(v^2)=2\frac{dv}{dt}v$. We can rewrite our integral as \begin{align} \int_0^tdt'\frac{dv}{dt'}v(t')&=\int_0^tdt'\frac{d}{dt'}\left(\frac 1 2v(t')^2\right)\\ &=\left[\frac 1 2v(t')^2\right]_0^t\\ &=\frac 1 2v(t)^2-\frac 1 2v(0)^2 \end{align} If we plug this back into our original expression we will still get something weird: \begin{align} \frac{d}{dt}W&=\frac{d}{dt}\left(\frac 1 2mv(t)^2-\underbrace{\frac 1 2mv(0)^2}_{\text{constant}}\right)\\ &=\frac{d}{dt}\left(\frac 1 2mv(t)^2\right) \end{align} Why do we get this weird expression? We want to know the average power, not the power as a function of $t$. We have to write $$P_\text{avg}=\frac{\Delta W}{\Delta t}=\frac{W(t)}{t},$$ like in method 1. In the above derivation we have shown that $$W(t)=\left(\frac 1 2mv(t)^2-{\frac 1 2mv(0)^2}\right)$$ so this is consistent with what we should expect.

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  • $\begingroup$ Very well explained, btw I love your username $\endgroup$
    – Numenorean
    Jun 5 at 16:35
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Your second approach is wrong because $\vec v$ is definitely not constant and $\vec F$ also may not be constant. Note that the question is rather ambiguous, it does not clearly state if instantaneous or average power is requested and it does not state if the force or power are constant.

The first part calculates the average power. In the second part $\vec F \cdot \vec v$ is the instantaneous power. So they are fundamentally different quantities. The average power does not depend on $\vec F$ being constant, but the instantaneous power does.

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The question is poorly worded. The setter almost certainly wants you to calculate the mean power. That is what you have done correctly (except for the missing unit!) in your first procedure.

The instantaneous power is $Fv$. The derivation is very simple ... $$\text {Power}= \lim_{\Delta t \to 0} \frac {\text{Work done in time}\ \Delta t}{\Delta t}=\lim_{\Delta t \to 0} \frac{F\Delta x}{\Delta t}=Fv$$

Therefore, assuming a constant force, the power increases from zero to a maximum value over the 8 s interval because $v$ continuously increases. Your second procedure gives the power just before the end of the 8 s, making the same assumption. It gives this 'final' power because you have substituted the final velocity, 25, for $v$.

Not surprisingly, you have found the final power to be twice the mean power!

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