Question

A 1 kg block situated on a rough incline is connected to a spring of spring constant 100Nm−1100Nm−1 as shown. The block is released from rest with the spring in the unstretched position. The block moves $10cm$ 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

At equilibrium
Normal reaction, $R=mg\cos {37}^{\circ }$
Frictional force , $f=\mu R=mg\sin {37}^{\circ }$
Where , $\mu$ is the coefficient of friction
Net force acting on the block $=mg\sin {37}^{\circ }-f$
$=mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ }$
$=mg(\sin {37}^{\circ }-\mu \cos {37}^{\circ })$
At equilibrium the work done by the block is equal to the potential energy of the spring i.e.,
$mg(\sin {37}^{\circ }-\mu \cos {37}^{\circ })x=\left(1/2\right)k{x}^2$
$1\times 9.8(\sin {37}^{\circ }-\mu \cos {37}^{\circ })=\left(1/2\right)\times 100\times \left(0.1\right)$
$0.602-\mu \times 0.799=0.510$
$\therefore \mu =0.092/0.799=0.115$