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(A) For a smooth inclined plane:The acceleration is $a_1 = g \sin 30^{\circ} = \frac{g}{2}$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,wi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) For a smooth inclined plane:The acceleration is $a_1 = g \sin 30^{\circ} = \frac{g}{2}$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,with $u = 0$:$S = \frac{1}{2} \left(\frac{g}{2}\right) T^2 = \frac{g T^2}{4} \quad \dots (i)$For a rough inclined plane:The acceleration is $a_2 = g \sin 30^{\circ} - \mu g \cos 30^{\circ} = g(\sin 30^{\circ} - \mu \cos 30^{\circ}) = g\left(\frac{1}{2} - \mu \frac{\sqrt{3}}{2}\right) = \frac{g}{2}(1 - \sqrt{3}\mu)$.Using the same distance $S$ and time $\alpha T$:$S = \frac{1}{2} \left[\frac{g}{2}(1 - \sqrt{3}\mu)\right] (\alpha T)^2 = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2 \quad \dots (ii)$Equating $(i)$ and $(ii)$:$\frac{g T^2}{4} = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2$$1 = (1 - \sqrt{3}\mu) \alpha^2$$\frac{1}{\alpha^2} = 1 - \sqrt{3}\mu$$\sqrt{3}\mu = 1 - \frac{1}{\alpha^2} = \frac{\alpha^2 - 1}{\alpha^2}$$\mu = \frac{1}{\sqrt{3}} \left(\frac{\alpha^2 - 1}{\alpha^2}\right)$Comparing this with the given expression $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$,we get $x = 3$.

Similar Questions

If an inclined plane is made slowly horizontal by reducing its inclination with the horizontal,the component of weight parallel to the plane of a block resting on the inclined plane:

$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

For the equilibrium of a body on an inclined plane of inclination $45^{\circ}$, the coefficient of static friction must be:

If for an inclined plane the coefficient of static friction is $\mu_s = \frac{3}{4}$,then the angle of repose for the inclined plane will be ........ $^o$.

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