A box initially at rest is on a plank, of length 5m, that is elevated at an angle such that tan(a)=3/4. When it reaches the end of the plank it has velocity 5ms^-1. Calculate the average frictional force on the box.
In physics questions it is useful to not plug numbers in until the end and instead work through algebraically. Let's say the angle of elevation is a, the mass is m, the velocity at the end of the plank is v and the length is L.Going down the plank the box converts its potential energy into kinetic energy. The discrepancy between the change in potential energy and the change in the kinetic energy will be due to energy loss due to friction.The change in potential is U=mgLsin(a)The change in kinetic energy is K=1/2 mv2ELoss= U-K = mgLsin(a)-1/2 mv2 The frictional force will be roughly constant across the plank as the normal contact force is constant, therefore ELoss+FFrictionxL=> FFriction= m( gsin(a) - 1/2 mv2/L ) = 8.5N
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