A 5.80-kg block is set into motion up an inclined plane with an initial speed of

Question

A 5.80-kg block is set into motion up an inclined plane with an initial speed of vi8.20 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 = 30.0° 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 KE = 5.80 ( 0^2 - 8.20^2) /2

= - 195 J

(b) change in gPE = m g d sin30 = 85.26 J

(C) applying work-energy theorem,

Work done by gravity + work done by friction = change in KE

- 85.26 - (3f) = - 195

f = 36.6 N

(d) N = m g cos30 = 49.22

u = f / N = 0.74

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