Question

A $1kg$ block situated on a rough inclined plane is connected to a spring of spring constant $100N{m}^{-1}$ as shown in figure. The block of released from rest with the spring in the unstretched position. The block moves $10cm$ along the incline before coming to rest. Find the coefficient of friction between the block and the incline assume that the spring has negligible mass and the pulley is frictionless. Take $g=10m{s}^{-2}$.

Answer 1

At the equilibrium the normal reaction is given as,

$N=mg\cos {37}^{\circ }$

The frictional force is given as,

$f=\mu N$

$=\mu mg\cos {37}^{\circ }$

The net force acting on the block is given as,

$F=mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ }$

Since, the work done by the block will be equal to the potential energy of the spring.

It can be written as,

$\left(mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ }\right)\times x=\frac{1}{2}k{x}^2$

$1\times 10\times \left(\sin {37}^{\circ }-\mu \cos {37}^{\circ }\right)\times \left(0.1\right)=\frac{1}{2}\times 100{\left(0.1\right)}^2$

$\mu =0.127$

Thus, the coefficient of friction between the block and the incline is $0.127$.