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(C) The net force acting on the body along the inclined plane is given by $F_{net} = mg \sin 	heta - f_k$,where $f_k = \mu N = \mu mg \cos 	heta$.Si

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at $x = 2.5 \, m$

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(C) The net force acting on the body along the inclined plane is given by $F_{net} = mg \sin 	heta - f_k$,where $f_k = \mu N = \mu mg \cos 	heta$.Since $\mu = 0.3x$,the net force is $F_{net} = mg \sin 	heta - (0.3x) mg \cos 	heta$.The acceleration of the body is $a = \frac{F_{net}}{m} = g \sin 	heta - 0.3x g \cos 	heta$.The speed of the body increases as long as the acceleration is positive. The speed is maximum when the acceleration becomes zero.Setting $a = 0$:$g \sin 37^{\circ} - 0.3x g \cos 37^{\circ} = 0$Given $\sin 37^{\circ} = 0.6$ and $\cos 37^{\circ} = 0.8$:$10 	imes 0.6 - 0.3 	imes x 	imes 10 	imes 0.8 = 0$$6 - 2.4x = 0$$2.4x = 6$$x = \frac{6}{2.4} = 2.5 \, m$.

$A$ body is moving down a long inclined plane of slope $37^{\circ}$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed at:

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