# Question of force of friction on incline plane [closed]

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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 Welcome to Physics! Generally we discourage questions that just ask for someone to check your work. Once you have identified the specific concept that you're not sure about, that's the point at which it's appropriate to ask a question here. If you can edit this question to be a specific conceptual question, I'll be happy to reopen it. – David Zaslavsky♦ Mar 6 '12 at 23:04

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

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