A 1kg block situated on a rough incline is connected to a spring … - Vidyadip

Question

A $1kg$ block situated on a rough incline is connected to a spring of spring constant $100Nm^{-1}$ as shown in. 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.

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Answer

Mass of the block, $m = 1kg$ Spring constant, $k = 100Nm^{-1}$ Displacement in the block, $X = 10cm = 0.1m$ The given situation can be shown as in the following figure.At equilibrium:

Normal reaction, $\text{R}=\text{mg}\cos37^{\circ}$ Frictional force, $\text{f}=\mu\text{R}=\text{mg}\sin37^{\circ}$ Where, $\mu$ is the coefficient of friction Net force acting on the block $ =\text{mg}\sin 37^{\circ}-\text{f}$ $=\text{mg}\sin37^{\circ}-\mu\text{m}\cos37^{\circ}$ $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})$ At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e., $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})\text{x}=\Big(\frac{1}{2}\Big)\text{kx}^2$ $1\times9.8(\sin37^{\circ}-\mu\cos37^{\circ})=\Big(\frac{1}{2}\Big)\times100\times(0.1)$ $0.602-\mu\times0.799=0.510$ $\therefore\mu=\frac{0.092}{0.799}=0.115$

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