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An object of mass $4\text{ kg}$ starts at rest from the top of a rough inclined plane of height $10\text{ m}$. If the speed of the object at the bottom of the inclined plane is $10\text{ m/s}$, and letting $g=10\ \mathrm{m/s^2}$ how much work is done by the force of friction?

This is as far as I got: if the angle of the incline is $\theta$, and $d$ is the length of the hypotenuse of the incline, then $\sin(\theta)=10/d$ so $d=10/\sin(\theta)$. Decomposing the mass of the object and projecting onto the axis of the friction force, $f$, I get $\sin(\theta)mg=f$. Now I can plug into the first equation to get $d=10/\sin(\theta)=10/(f/mg)$.

$$W=Fd=f(10/\sin(\theta))=f(10/(f/mg))=10mg=4\times 10\times 10=400\text{ J}$$ but my notes say the answer is $200\text{ J}$.

What am I doing wrong?

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closed as too localized by David Zaslavsky♦ Mar 6 '12 at 23:05

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1 Answer

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The problem is with this approach is that W = F x d assumes a constant force. Your frictional force isn't constant, since you start from rest (static friction) and end up sliding (dynamic friction). Also, one clue that this approach may not be entirely correct is the fact that you didn't use the velocity provided (though this could be a red herring).

Try working the problem out from an energy standpoint, instead of using forces. Energy is often a preferable method to forces from an ease-of-calculation standpoint.

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I was thinking of using conservation of mechanical energy, but that doesn't get me anywhere because I know all the variables. – jay Mar 6 '12 at 23:05
@jay Not true. You don't know the one you're solving for, i.e. the work done by (energy lost to) friction. – Mitchell Mar 6 '12 at 23:09
Mitchell I got!!! Thanks so much, you made me realize this is work done by the nonconservative force of friction – jay Mar 6 '12 at 23:18
@jay That it is. Glad to be of service. – Mitchell Mar 6 '12 at 23:27

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