Forces-Work question

This is functionally equivalent to asking the speed of a box of mass m sliding with friction coeff u_k down a slope at angle theta after descending a vertical distance h.

Is the acceleration constant? If the drop were vertical, the acceleration would be constant at a value of g, given by a force = m.g But there is friction, so the force is reduced, but still constant. What is it? Downslope force = m.g sin theta - u_k . m . cos theta So downslope acceleration is force / mass = g . sin theta - u_k . cos theta (which is constant)

The usual equation of constant acceleration, given velocity distance and acceleration is v^2 = u^2 + 2 a x We have a value for u (zero starting speed) accel a (given above) x = downslope distance

Only x is now unknown. It is h / sin theta

Putting this stuff together gives v^2 = 2 (g sin theta - u_k cos theta) ( h/sin theta ) = 2h ( g - u_k cot theta) So v = sqrt ( 2 g h - 2 u_k h cot theta )

(If this is homework, you may be unpleasantly surprised if I left a mistake in this - I wrote it off the top!)

Brian Whatcott Altus, OK

Use energy....

1/2 m v^2 ( start ) = mgh ( end ) + ( h/ (sin theta)) u (cos theta) mg ( friction loss )

note m cancels out , cos / sin = 1/tan

1/2 v^2 = gh + hug/tan theta

Thanks for the replies guys, just one last thing:

Jonathan Barnes got his answer as:

1/2 v^2 = gh + hug/tan theta

whilst Brian Whatcott got his answer as:

1/2 v^2 = gh - hu/tan(theta)

There is a missing g in the second answer, I think it may have come from here: Downslope force = m.g sin theta - u_k . m . cos theta

Im thinking it should be Downslope force = m.g sin theta - u_k . m . g. cos theta

Am I correct?

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Yes

Brian W