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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A wall is tilted in by $15^{\large\circ}$,and a 2kg block is placed against the wall.The block is held in place with a spring.The coefficients of friction between the block and the wall are $\mu_s=0.61$ and $\mu_k=0.49$.Holding the block in place requires that the spring be compressed by 12cm.What is the spring constant of the spring?

$\begin{array}{1 1}301N/m\\300N/m\\201N/m\\200N/m\end{array} $

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Answer : 301N/m
$-N+kx-mg\sin 15^{\large\circ}+\mu_s N-mg\cos 15^{\large\circ}=0$
$N=\large\frac{mg\cos 15^{\Large\circ}}{\mu_s}$
$kx =mg\sin 15^{\large\circ}+\large\frac{mg\cos 15^{\Large\circ}}{\mu_s}=$$mg\big(\sin 15^{\large\circ}+\large\frac{\cos 15^{\large\circ}}{\mu_s}\big)$
$k=\large\frac{mg}{x}$$\big(\sin 15^{\large\circ}+\large\frac{\cos 15^{\large\circ}}{\mu_s}\big)$
$\;\;\;=\large\frac{(2kg)(9.8m/s^2)}{0.12m}$$\big(\sin 15^{\large\circ}+\large\frac{\cos 15^{\Large\circ}}{0.61}\big)$
$\;\;\;=301N/m$
answered Sep 2, 2014 by sreemathi.v
 

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