A 5.60-kg block is set into motion up an inclined plane with an initial speed of

Question

A 5.60-kg block is set into motion up an inclined plane with an initial speed of vi = 7.60 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 -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

Given,

vi = 7.6 m/s ; m = 5.6 kg ; d = 3 m ; theta = 30 deg ; vf = 0

a)The KE change will be:

delta-KE = 1/2 m (vf^2 - vi^2)

delta-KE = 0.5 x 5.6 (0^2 - 7.6^2) = -161.73 J

Hence, delta-KE = -161.73 J

b)delta-PE = m g d sin(theta)

delta-PE = 5.6 x 9.81 x 3 x sin30 = 82.40

delta-PE = 82.4 J

c)The work done against frictional force is:

W = 161.73 - 82.4 = 79.33 J

W = Ff d

Ff = W/d = 79.33/3 = 26.44 N

Hence, Ff = 26.44 N

d)We know that

Ff = u m g cos(theta)

u = Ff/mg cos(theta)

u = 26.44/(5.6 x 9.81 x cos30) = 0.56

hence, u = 0.56

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