Question

A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m¯¹ as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer 1

Given, the mass of the block is 1 kg, spring constant is 100 N/m and the distance moving down the incline by the block is 10 cm or 0.1 m.

The following diagram shows the forces acting on the block while moving down the inclined plane.

Here, f is the frictional force acting on the block and R is the normal reaction force.

The expression for the frictional force acting on the block is,

f=μR……. (1)

Here, μ is the coefficient of friction.

Balance the forces vertically with respect to the block as the block is in equilibrium with respect to vertical position of block,

R=mgcos37°……. (2)

Let net force F acts on the block horizontally, then

F=mgsin37°−f

From equation (1) and (2),

F=mgsin37°−μ( mgcos37° )

When the block moves some distance and then comes to rest, the work done by the block in moving down is stored in the spring in the form of potential energy, hence

Fd= 1 2 k d 2

Here, d is the distance moved by the block and k is the spring constant.

Substitute the values in the above expression.

( mgsin37°−μmgcos37° )( 0.1 )= 1 2 ×100× ( 0.1 ) 2 1×10( sin37°−μcos37° )=0.51 μ=0.125

Hence, the value of coefficient of friction is 0.125.