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A $1 \mathrm{~kg}$ block situated on a rough incline is connected to a spring of spring constant $100 \mathrm{~N} / \mathrm{s}$ as shown in the fig. The block is released from rest with the spring in the unstretched position. The block moves $10 \mathrm{~cm}$ down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring is of negligible mass and the pulley is frictionless.
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From figure, $\mathrm{R}=\mathrm{mg} \cos \theta$ and $\mathrm{F}=\mu \mathrm{r}=\mu \mathrm{mg} \cos \theta$
Net force on the block in the downward direction
$=\mathrm{mg} \sin \theta-\mathrm{F}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta$
$=m g(\sin \theta-\mu \cos \theta)$
Distance moved $=x=10 \mathrm{~cm}=0.1 \mathrm{~m}$
In equilibrium, work done $=$ PE of the spring
$=\mathrm{mg}(\sin \theta-\mu \cos \theta) x=\frac{1}{2} k x^2$
$2 \mathrm{mg}(\sin \theta-\mu \cos \theta)=\mathrm{k} x$
$2 \times 1 \times 10\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)=100 \times 0.1$
$20(0.601-\mu \times 0.798)=10 \Rightarrow \mu=0.126$
Net force on the block in the downward direction
$=\mathrm{mg} \sin \theta-\mathrm{F}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta$
$=m g(\sin \theta-\mu \cos \theta)$
Distance moved $=x=10 \mathrm{~cm}=0.1 \mathrm{~m}$
In equilibrium, work done $=$ PE of the spring
$=\mathrm{mg}(\sin \theta-\mu \cos \theta) x=\frac{1}{2} k x^2$
$2 \mathrm{mg}(\sin \theta-\mu \cos \theta)=\mathrm{k} x$
$2 \times 1 \times 10\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)=100 \times 0.1$
$20(0.601-\mu \times 0.798)=10 \Rightarrow \mu=0.126$
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