Question

A spring block system is on a horizontal surface. The block is compressed by 10cm. What should be the coefficient of friction between the block and ground such that the block comes to a stop at the mean position? Mass of block = 0.5kg & spring constant = 3N/m.

(a) 0.03

(b) 0.06

(c) 0.12

(d) 0.05

Answer 1

The correct choice is (b) 0.06

The explanation: When the block is released, the spring force will act towards the mean position and friction force will act against velocity. The work done by these forces should be equal to potential energy of the system in the compressed state.                                                                                                                                                   ∴ Work done in dx displacement = (μmg – kx)dx.

On integrating this, with dx varying from 0 to 0.1, we get   

Work = (μmgx – kx√/2)0^0.1= 0.5μ – 0.015.

This should be equal to initial potential energy of the system.     

0.5μ – 0.015 = 1/2kA√

∴ 0.5μ  – 0.015 = 1/2 * 3 * 0.01

∴ 0.5μ  – 0.015 = 0.015            

∴ 0.5μ = 0.03 OR μ = 0.06.