Question

The time taken by a particle to move down an inclined plane of inclination angle alpha is T1, and to move down another inclined plane of same height, but with inclination angle beta is t2. Of the ratio T1 /t2 = √3, and alpha = 60°, then beta = ? (A)60° (B)30° (C)15° (D)75°

Answer 1

angle of incline = θ.

Initially it is a smooth surface.

Let the mass be m. Gravity g. 

Initial velocity = u = 0.

acceleration down the incline = a = g sinθ

time taken = T.

So length of incline = s = 1/2 * g sinθ * T²   ---(1)

When there is dynamic friction of coefficient = μ,

    friction force = f = μ N = μm g Cos θ

a = (sin θ - μ Cos θ) g  

Time taken is = 2 T

So we get:   s = 1/2 * g (sinθ - μ Cos θ) * (2T)²    ---(2)

From (1) and (2) , we get

        Sin θ = (sinθ - μ cosθ) 4

        μ = Tanθ  * 3/4

Answer 2

Answer:

angle of incline = θ.

Initially it is a smooth surface.

Let the mass be m. Gravity g. 

Initial velocity = u = 0.

acceleration down the incline = a = g sinθ

time taken = T.

So length of incline = s = 1/2 * g sinθ * T²   ---(1)

When there is dynamic friction of coefficient = μ,

    friction force = f = μ N = μm g Cos θ

a = (sin θ - μ Cos θ) g  

Time taken is = 2 T

So we get:   s = 1/2 * g (sinθ - μ Cos θ) * (2T)²    ---(2)

From (1) and (2) , we get

        Sin θ = (sinθ - μ cosθ) 4

        μ = Tanθ  * 3/4