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(B) According to the Work-Energy Theorem,the total work done on the bar is equal to the change in its kinetic energy.Since the bar starts from rest an

Vedclass · Accepted Answer

$\frac{2 	an 	heta}{k}$

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(B) According to the Work-Energy Theorem,the total work done on the bar is equal to the change in its kinetic energy.Since the bar starts from rest and stops at distance $S$,the change in kinetic energy $\Delta K = 0 - 0 = 0$.Therefore,the total work done $W = \int_{0}^{S} (mg \sin 	heta - \mu mg \cos 	heta) dx = 0$.Substituting $\mu = kx$,we get $\int_{0}^{S} (mg \sin 	heta - kx mg \cos 	heta) dx = 0$.Integrating with respect to $x$: $[mg \sin 	heta \cdot x - \frac{1}{2} k mg \cos 	heta \cdot x^2]_0^S = 0$.This simplifies to $mg \sin 	heta \cdot S = \frac{1}{2} k mg \cos 	heta \cdot S^2$.Dividing both sides by $mg \cos 	heta \cdot S$ (assuming $S 
eq 0$),we get $	an 	heta = \frac{1}{2} k S$.Solving for $S$,we find $S = \frac{2 	an 	heta}{k}$.

$A$ small bar starts sliding down an inclined plane forming an angle $\theta$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$,where $k$ is a constant. Find the distance covered by the bar until it stops.

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