A block of mass 2,,kg is placed on a rough in | vedclass

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(D) Given: Mass $m = 2\,\,kg$,coefficient of friction $\mu = 0.2$,angle of inclination $	heta = 30^{\circ}$,spring constant $k = 100\,\,N/m$,accelera

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$13\,\,cm$

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(D) Given: Mass $m = 2\,\,kg$,coefficient of friction $\mu = 0.2$,angle of inclination $	heta = 30^{\circ}$,spring constant $k = 100\,\,N/m$,acceleration due to gravity $g = 10\,\,m/s^2$.According to the Work-Energy Theorem,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.Work done by gravity $(W_g)$ + Work done by friction $(W_f)$ + Work done by the spring $(W_s)$ = $0$.$W_g = mg \sin(30^{\circ}) \cdot x$$W_f = -\mu mg \cos(30^{\circ}) \cdot x$$W_s = -\frac{1}{2} kx^2$Substituting the values:$mg \sin(30^{\circ})x - \mu mg \cos(30^{\circ})x - \frac{1}{2} kx^2 = 0$Dividing by $x$ (assuming $x 
eq 0$):$mg \sin(30^{\circ}) - \mu mg \cos(30^{\circ}) - \frac{1}{2} kx = 0$$(2)(10)(0.5) - (0.2)(2)(10)(0.866) = \frac{1}{2}(100)x$$10 - 3.464 = 50x$$6.536 = 50x$$x = \frac{6.536}{50} = 0.1307\,\,m = 13.07\,\,cm \approx 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

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