A small block starts sliding down an inclined | vedclass

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Question

(A) Given that, $\mu = C s^2$.The net force on the block along the incline is given by:$M g \sin 	heta - f = M a$$M g \sin 	heta - \mu M g \cos 	he

Vedclass · Accepted Answer

$\sqrt{\frac{3}{C}}$

Vedclass · Answer

(A) Given that, $\mu = C s^2$.The net force on the block along the incline is given by:$M g \sin 	heta - f = M a$$M g \sin 	heta - \mu M g \cos 	heta = M a$$a = g \sin 	heta - \mu g \cos 	heta$Substituting $\mu = C s^2$ and $	heta = 45^{\circ}$:$a = g \sin 45^{\circ} - C s^2 g \cos 45^{\circ} = \frac{g}{\sqrt{2}} (1 - C s^2)$Using the relation $a = v \frac{dv}{ds}$, we have:$v \frac{dv}{ds} = \frac{g}{\sqrt{2}} (1 - C s^2)$$v dv = \frac{g}{\sqrt{2}} (1 - C s^2) ds$Integrating both sides from initial state $(s=0, v=0)$ to final state $(s=s_{max}, v=0)$:$\int_{0}^{0} v dv = \int_{0}^{s_{max}} \frac{g}{\sqrt{2}} (1 - C s^2) ds$$0 = \frac{g}{\sqrt{2}} [s - \frac{C s^3}{3}]_{0}^{s_{max}}$Since $g 
eq 0$, we have:$s_{max} - \frac{C s_{max}^3}{3} = 0$$s_{max} (1 - \frac{C s_{max}^2}{3}) = 0$Since $s_{max} 
eq 0$, we get:$1 = \frac{C s_{max}^2}{3}$$s_{max}^2 = \frac{3}{C}$$s_{max} = \sqrt{\frac{3}{C}}$

$A$ small block starts sliding down an inclined plane forming an angle $45^{\circ}$ with the horizontal. The coefficient of friction $\mu$ varies with distance $s$ as $\mu = C s^2$, where $C$ is a constant of appropriate dimensions. The distance covered by the block before it stops is:

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