a 10.0 kg block slides down a plane inclined at an angle of 53 degrees to the ho

Question

a 10.0 kg block slides down a plane inclined at an angle of 53 degrees to the horizontal from an additional height of 40.0 m the coefficient of kinetic friction between the block and the inclined plane is 0.300. the block continues to slide 20.0 m across a horizontal surface. the coefficient of kinetic friction between the block andf the horizontal surface is 0.200. the block then continues up a frictionless ramp inclined at an angle theta,, to the horizontal determine the total work done agaist friction and the final height the block will rise.

Explanation / Answer

fk - mg sin 53 = ma
k N - mg sin53 = ma    ( N =mgcos30) (be very careful here)
0.30 ( mgcos53) - mg sin53= ma
a = -6.058m/s2 (the negative sign meansdecreasing in acceleration)

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