Consider a small block sliding down an inclin | vedclass

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(B) The normal force on the block is $N = mg \cos 	heta$.Since the coefficient of friction is $\mu = \frac{2}{3} x$, the kinetic friction force is $f

Vedclass · Accepted Answer

$\sqrt{3} 	ext{ m}$

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(B) The normal force on the block is $N = mg \cos 	heta$.Since the coefficient of friction is $\mu = \frac{2}{3} x$, the kinetic friction force is $f_k = \mu N = \left( \frac{2}{3} x ight) mg \cos 	heta$.According to the work-energy theorem, the total work done on the block is equal to the change in its kinetic energy: $W_g + W_f = \Delta K.E. = 0 - 0 = 0$.The work done by gravity as the block slides down a distance $x$ is $W_g = mgx \sin 	heta$.The work done by the variable friction force is $W_f = - \int_0^x f_k \, dx = - \int_0^x \left( \frac{2}{3} x mg \cos 	heta ight) dx = - \frac{2}{3} mg \cos 	heta \left[ \frac{x^2}{2} ight]_0^x = - \frac{1}{3} mgx^2 \cos 	heta$.Substituting these into the work-energy equation: $mgx \sin 	heta - \frac{1}{3} mgx^2 \cos 	heta = 0$.Dividing by $mgx$ (assuming $x 
eq 0$): $\sin 	heta - \frac{1}{3} x \cos 	heta = 0$.Solving for $x$: $x = 3 	an 	heta$.Given $	heta = 30^{\circ}$, $x = 3 	an 30^{\circ} = 3 	imes \frac{1}{\sqrt{3}} = \sqrt{3} 	ext{ m}$.

Consider a small block sliding down an inclined plane of inclination $30^{\circ}$ with the horizontal. The coefficient of friction is $\mu = \frac{2}{3} x$, where $x$ is the distance (in meters) through which the mass slides down. The distance covered by the mass before it stops is

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