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A car ($m=540\,\text{kg}$) engine, has a power of $60\,\text{kW}$. The static friction coefficient between wheels and road is $k=0.6$. How long does it take to reach the speed of $27.7\,\text{m/s}$, with constant acceleration?

I have tried the following:

The energy to reach given speed is: $$\frac{mv^2}{2}=\frac{540\cdot 27.7^2}{2}=2.07\cdot 10^5\,\text{J}$$

In the meanwhile I dissipated (because of friction): $$-F_a\cdot s=-mgk\cdot\frac{1}{2}at^2=-mgk\cdot\frac{1}{2}\frac{\Delta v}{t}t^2=158,922\cdot t$$

The engine can do a work of $60,000\,\text{J}$ per second, so the work done is $60,000\cdot t$

So, I can do: $$2.07\cdot 10^5\,\text{J}+158,922\cdot t=60,000\cdot t$$ And I obtain a negative time. How is possible? What's wrong? Thanks a lot

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You should check your working for the energy dissipated due to friction. I get $43,976.52W$. –  Jason Davies Dec 5 '12 at 11:36
@JasonDavies I have checked and checked again! Could you highlight my error? Thanks again –  Surfer on the fall Dec 5 '12 at 11:38
I just put the numbers in. Not sure what else I can say. Maybe you should show what numbers you used? –  Jason Davies Dec 5 '12 at 11:46
Working backwards, it looks like you forgot to put the number in for $\Delta{v}$. –  Jason Davies Dec 5 '12 at 11:51
@JasonDavies OMG, you're absolutely right. In my textbook there was $v_f=100 km/h$ originally, which I converted to $27,7 m/s$. When I was solving the problem, I forgot the conversion and I have used $\Delta v=100$. You saved my day. Thanks a lot! –  Surfer on the fall Dec 5 '12 at 11:53

2 Answers 2

up vote 1 down vote accepted

Where you went wrong was $$-mgk \times \frac{1}{2}at^2=-mgk \times \frac{1}{2}\frac{\Delta v}{t}t^2$$. Instead it should be $$-mgk \times \frac{1}{2}\frac{\Delta v}{\Delta t}t^2$$ Anyways I would do it as $$F_{friction}s = mgk\frac{1}{2}at^2 = \frac{v-u}{2t}mgkt^2 = \frac{v-u}{2}mgkt = \frac{27.7ms^{-1}}{2} \times 540kg \times {10ms^{-2}} \times 0.6 \times t = (44874kgm^2s^{-3})t$$ So $$60kW \times t = (44874kgm^2s^{-3})t + 207000J$$ and solving gives $$t=13.7s$$

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Wait... I seem to have got it wrong. Wait for me to rethink this thing... –  namehere Dec 5 '12 at 11:42
No problem. The result reported on my textbook is 13s. Thanks a lot! –  Surfer on the fall Dec 5 '12 at 11:46
I'll change it later to account for the fact that u is 100 meters per second –  namehere Dec 5 '12 at 12:06
Darn, no matter which way I put it I couldn't get the answer. Sorry for that. I'm giving up. –  namehere Dec 5 '12 at 12:44
@Surferonthefall Eureka! I'll edit my answer now! –  namehere Dec 5 '12 at 12:47

If you have worked out it correctly then this simply means that you cant reach that speed with that powerful engine while the frictional coefficient is that much.What if you add some positive number to one side of your last equation so that time will now be positive?Then the left and right side of your equation will be same for some positive time but what does it mean of adding positive number to one side of equation?As you will add the number but it should have the same unit as of energy,then actually you are adding additional energy to the car engine.So without that much adding of additional energy you cant reach that velocity magnitude.

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