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.

Answer 1

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,
μ_kR + ma − mg sin θ = 0
(where μ_k is the coefficient of kinetic friction)
From Equation (1):
μ_kmg cos θ + ma − mg sin θ = 0
⇒ m (μ_kg cos θ + a − g sin θ) = 0
⇒ μ_k × 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.