A 5.00 kg block is set into motion up an inclined plane withan initial speed of 8.00 m/s. The block comes to rest aftertraveling 3.00 m along the plane, which is inclined at an angle of30.0 degrees to the horizontal. For this motion determine: (a) thechange in the block's kinetic energy, (b) the change in potentialenergy, (c) thefriction force exerted on the block (assume it to be constant), (d)What is the coefficient of kinetic friction?
a) The change in kinetic energy is: . (Ignore the boxes) Since the finalvelocity is zero the change in KE is There should be a minus sign infront of the KE expression.
The change in potential energy is equal to the negative of the workdone by gravity which is:
c) i have no clue how they got 28.8 for the force offriction?
d) the answer is 0.679


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