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(A) The acceleration of the mass is given by $a = g(\sin 	heta - \mu_0 x \cos 	heta)$.Using the work-energy theorem or integrating $v dv = a dx$,we

Vedclass · Accepted Answer

$\sqrt{\frac{g 	an 	heta \sin 	heta}{\mu_0}}$

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(A) The acceleration of the mass is given by $a = g(\sin 	heta - \mu_0 x \cos 	heta)$.Using the work-energy theorem or integrating $v dv = a dx$,we have $\int_0^{v} v dv = \int_0^{x/2} g(\sin 	heta - \mu_0 x \cos 	heta) dx$.Integrating both sides: $\frac{v^2}{2} = g[\sin 	heta \cdot x' - \frac{\mu_0}{2} (x')^2 \cos 	heta]$,where $x' = x/2$.Substituting $x' = \frac{x}{2} = \frac{1}{\mu_0} 	an 	heta$ (since maximum distance $x_{max} = \frac{2 	an 	heta}{\mu_0}$):$\frac{v^2}{2} = g[\sin 	heta (\frac{	an 	heta}{\mu_0}) - \frac{\mu_0}{2} (\frac{	an 	heta}{\mu_0})^2 \cos 	heta]$.$\frac{v^2}{2} = g[\frac{\sin 	heta 	an 	heta}{\mu_0} - \frac{	an^2 	heta \cos 	heta}{2 \mu_0}] = g[\frac{\sin 	heta 	an 	heta}{\mu_0} - \frac{\sin 	heta 	an 	heta}{2 \mu_0}] = \frac{g \sin 	heta 	an 	heta}{2 \mu_0}$.Thus,$v^2 = \frac{g \sin 	heta 	an 	heta}{\mu_0}$,which gives $v = \sqrt{\frac{g 	an 	heta \sin 	heta}{\mu_0}}$.

In the above question,the speed of the mass when it has traveled half the maximum distance is:

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