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A $1\; kg$ block situated on a rough incline is connected to a spring of spring constant $100\;N m ^{-1}$ as shown in Figure. 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.
$0.564$
$0.368$
$0.115$
$0.256$
Solution
Mass of the block, $m=1 kg$
Spring constant, $k=100 N m ^{-1}$
Displacement in the block, $x=10 cm =0.1 m$
The given situation can be shown as in the following figure.
At equilibrium:
Normal reaction, $R=m g \cos 37^{\circ}$
Frictional force, $f{=} \mu_{R}=m g \sin 37^{\circ}$
Where, $\mu$ is the coefficient of friction
Net force acting on the block $=m g \sin 37^{\circ}-f$
$=m g \sin 37^{\circ}-\mu m g \cos 37^{\circ}$
$=m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)$
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
$m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right) x=\frac{1}{2} k x^{2}$
$1 \times 9.8\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)=\frac{1}{2} \times 100 \times 0.1$
$0.602-\mu \times 0.799=0.510$
$\therefore \mu=\frac{0.092}{0.799}=0.115$
