A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ = 30° from the horizontal. The block is originally located 5m from the bottom of the plane. If the block, undergoing constant acceleration down the ramp, slides to the bottom in 2 seconds, what is the coefficient of the kinetic friction μk between the block and the inclined plane?

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AFOKE88 AFOKE88 · Answer 1 · 2020-03-07T07:07:19+01:00

Answer:

Fk = 21.645N

Explanation:

Let Fb be Force of block and thus;

Fb= mg where m is mass of block and g is acceleration due to gravity

Thus Fb= 9kg x 9.81N/kg = 88.29 N

Now, the question says this force Fb rests at an inclined plane of@ 30° angle

Thus;

Force parallel to inclined plane = 88.29 x sin30° = 44.145 N.

Force perpendicular to the inclined plane = 88.29 x cos30 = 76.46 N

Now, when an object is falling freely, we know that

h = (1/2)at^(2)

From the question, the height is 5m and t= 2 seconds

Thus;

5 = (1/2)a(2)^(2)

2a = 5 and thus,

a = 5/2 = 2.5 m/s^(2)

Now, in inclined planes, perpendicular force - kinetic friction force = Resultant force

Thus let perpendicular be Fp and kinetic friction force be Fk and so;

Fp - Fk = F

F= ma = 9 x 2.5 = 22.5N

Thus, 44.145 - Fk = 22.5

Thus, Fk = 44.145 - 22.5 = 21.645N