Question

A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction µ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by µ = k tan θ. The value of k is ______.

विकल्प

• 3

• 1

• 2

• 4

Answer 1

A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction µ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by µ = k tan θ. The value of k is 3.

Explanation:

Given: The angle formed by the incline and the horizontal is, the block moves from point B to point A and comes to rest, the length BC is twice the length CA, the length BC of the incline is frictionless, and the length CA of the incline has friction with a coefficient of friction = k tan.

To find: The value of k.

Let the length AC = l, BC = 2l, AB = AC + BC = 3l

Applying Work-energy theorem:

Change in kinetic energy of the block = net work done on the block by gravity and friction

ΔK = mg(3l) sin θ - µmg(l) cos θ

0 = 3mgl sin θ - µmgl cos θ

µ = 3 tan θ

k = 3