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Calculate the force of friction (Ff) for a 5kg aluminum block being pulled with constant velocity (uniform motion), up a steel plank that is at an angle of 30 degrees to the horizontal.
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For constant velocity, the force pulling up the ramp must equal the forces pulling down the ramp, or else the block would be accelerating. There are two forces acting down the ramp: mgsin30 and Ff (frictional force), and one acting up the ramp. \[F _{f}= \mu _{k} mg \cos \theta \] Hmm. I think we need to know the coefficient of kinetic friction to work this out. Does it give you the coefficient of kinetic friction for steel?
If it doesn't, then \[F _{f}= \mu _{k} 5\times 9.8 \cos 30\]
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