A 4.40-kg block is set into motion up an inclined plane with an initial speed of

Question

A 4.40-kg block is set into motion up an inclined plane with an initial speed of v_i = 8.60 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of theta = 30.0degree to the horizontal. (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the block-Earth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

Explanation / Answer

a) change in kinetic energy = final kinetic energy - initial kinetic energy

= 0 - 0.5 x 4.4 x 8.62 = -162.712 J

b) change in potential energy = mgdsin30 = 4.4 x 9.8 x 3 x 0.5 = 64.68 J

c) friction force = (162.712 - 64.68)/3 = 32.677 N

d) u = 32.677/mgcos30 = 32.677/(4.4 x 9.8 x 0.866) = 0.875

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