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(A) The net force acting on the mass along the inclined plane is $F_{net} = mg \sin 	heta - f_r = ma$. Since the frictional force is $f_r = \mu N = \

Vedclass · Accepted Answer

$\frac{2}{\mu_0} 	an 	heta$

Vedclass · Answer

(A) The net force acting on the mass along the inclined plane is $F_{net} = mg \sin 	heta - f_r = ma$. Since the frictional force is $f_r = \mu N = \mu mg \cos 	heta$,we have $mg \sin 	heta - \mu mg \cos 	heta = ma$. Substituting $\mu = \mu_0 x$,the acceleration $a$ is given by $a = g(\sin 	heta - \mu_0 x \cos 	heta)$. Using the relation $a = v \frac{dv}{dx}$,we integrate the equation of motion: $\int_0^v v \, dv = \int_0^x g(\sin 	heta - \mu_0 x \cos 	heta) \, dx$. At the point where the mass stops,the final velocity $v = 0$. Thus,$0 = g [x \sin 	heta - \frac{\mu_0 x^2}{2} \cos 	heta]$. Setting the term inside the bracket to zero: $x \sin 	heta = \frac{\mu_0 x^2}{2} \cos 	heta$. Solving for $x$,we get $x = \frac{2 \sin 	heta}{\mu_0 \cos 	heta} = \frac{2}{\mu_0} 	an 	heta$.

$A$ small mass slides down an inclined plane of angle $\theta$ with the horizontal. The coefficient of friction is $\mu = \mu_0 x$,where $x$ is the distance through which the mass slides down and $\mu_0$ is a constant. Then,the distance covered by the mass before it stops is:

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