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Question
A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.
Solution
Free body diagram for the block is as follows:
From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ (1)
For the block, u = 0 m/s, s = 8 m and t = 2 s.
According to the equation of motion
`s=ut+1/2at^2`
`s=0+1/2a2^2`
8 = 2a
a = 4 m/s2
Again,
μkR + ma − mg sin θ = 0
(where μk is the coefficient of kinetic friction)
From Equation (1):
μkmg cos θ + ma − mg sin θ = 0
⇒ m (μkg cos θ + a − g sin θ) = 0
⇒ μk × 10 × cos 30° = g sin 30° − a
`=> mukxx10sqrt3/2=10xx(1/2)-4`
`=>(5sqrt3)muk=1)`
`=>muk=(1/(5sqrt3))=0.11`
Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.
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