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(D) According to the Work-Energy Theorem $(WET)$,the total work done on the block is equal to the change in its kinetic energy. Since the block starts

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$13$

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(D) According to the Work-Energy Theorem $(WET)$,the total work done on the block is equal to the change in its kinetic energy. Since the block starts from rest and comes to rest at maximum compression,the change in kinetic energy is zero.$W_{gravity} + W_{friction} + W_{spring} = 0$Let $x$ be the maximum compression of the spring.The work done by gravity is $W_g = mg \sin(30^{\circ}) \cdot x$.The work done by friction is $W_{f_r} = -\mu mg \cos(30^{\circ}) \cdot x$.The work done by the spring is $W_{spring} = -\frac{1}{2} k x^2$.Substituting these into the equation:$mg \sin(30^{\circ}) x - \mu mg \cos(30^{\circ}) x - \frac{1}{2} k x^2 = 0$$mg \sin(30^{\circ}) - \mu mg \cos(30^{\circ}) = \frac{1}{2} k x$Given: $m = 2 \, kg$,$g = 10 \, m/s^2$,$\mu = 0.2$,$k = 100 \, N/m$,$\sin(30^{\circ}) = 0.5$,$\cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$.$2 \cdot 10 \cdot 0.5 - 0.2 \cdot 2 \cdot 10 \cdot 0.866 = \frac{1}{2} \cdot 100 \cdot x$$10 - 3.464 = 50x$$6.536 = 50x$$x = \frac{6.536}{50} \approx 0.1307 \, m = 13.07 \, cm$.Rounding to the nearest integer,the compression is $13 \, cm$.

$A$ block of mass $2 \, kg$ is placed on a rough inclined plane as shown in the figure $(\mu = 0.2)$ so that it just touches the spring. The block is allowed to move downwards. The spring will be compressed to a maximum of .............. $cm$.

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