A 1; kg block situated on a rough incline is | vedclass

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Question

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 a

Vedclass · Accepted Answer

$0.115$

Vedclass · Answer

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 $=m g \sin 37^{\circ}-f$

$=m g \sin 37^{\circ}-\mu m g \cos 37^{\circ}$

$=m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}ight)$

At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,

$m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}ight) x=\frac{1}{2} k x^{2}$

$1 	imes 9.8\left(\sin 37^{\circ}-\mu \cos 37^{\circ}ight)=\frac{1}{2} 	imes 100 	imes 0.1$

$0.602-\mu 	imes 0.799=0.510$

$	herefore \mu=\frac{0.092}{0.799}=0.115$

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