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

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(B) Given: $m = 1 \,kg$, $k = 100 \,N \,m^{-1}$, $	heta = 45^{\circ}$, $x = 10 \,cm = 0.1 \,m$, $g = 10 \,m \,s^{-2}$.According to the work-energy th

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$0.3$

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(B) Given: $m = 1 \,kg$, $k = 100 \,N \,m^{-1}$, $	heta = 45^{\circ}$, $x = 10 \,cm = 0.1 \,m$, $g = 10 \,m \,s^{-2}$.According to the work-energy theorem, the work done by gravity minus the work done by friction equals the potential energy stored in the spring when the block comes to rest.Work done by gravity = $mg \sin 	heta \cdot x$Work done by friction = $f \cdot x = \mu N \cdot x = \mu mg \cos 	heta \cdot x$Potential energy of the spring = $\frac{1}{2} k x^2$Applying the work-energy principle:$mg \sin 	heta \cdot x - \mu mg \cos 	heta \cdot x = \frac{1}{2} k x^2$$mg \sin 	heta - \mu mg \cos 	heta = \frac{1}{2} k x$Substituting the values:$1 	imes 10 	imes \sin 45^{\circ} - \mu 	imes 1 	imes 10 	imes \cos 45^{\circ} = \frac{1}{2} 	imes 100 	imes 0.1$$10 	imes \frac{1}{\sqrt{2}} - \mu 	imes 10 	imes \frac{1}{\sqrt{2}} = 5$$\frac{10}{\sqrt{2}} (1 - \mu) = 5$$1 - \mu = \frac{5 \sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$\mu = 1 - \frac{1}{\sqrt{2}} \approx 1 - 0.707 = 0.293$Rounding to one decimal place, we get $\mu \approx 0.3$.

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