# Energy problem: What's wrong here?

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

• You should check your working for the energy dissipated due to friction. I get $43,976.52W$. Dec 5 '12 at 11:36
• @JasonDavies I have checked and checked again! Could you highlight my error? Thanks again 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? Dec 5 '12 at 11:46
• Working backwards, it looks like you forgot to put the number in for $\Delta{v}$. 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! Dec 5 '12 at 11:53

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 $$t=13.7s$$