A wedge of height H (fixed) and inclination a | vedclass

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(D) Let the wedge move with acceleration $g$ to the right. In the frame of the wedge, a pseudo force $F_p = mg$ acts on the block to the left.The comp

Vedclass · Accepted Answer

Time taken decreases with $\alpha$ for $0 < \alpha < \frac{\pi}{4}$.

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(D) Let the wedge move with acceleration $g$ to the right. In the frame of the wedge, a pseudo force $F_p = mg$ acts on the block to the left.The components of forces along the incline are:$1$. Component of gravity: $mg \sin \alpha$ (downwards along the incline).$2$. Component of pseudo force: $mg \cos \alpha$ (upwards along the incline).The net acceleration $a$ of the block up the incline is given by:$a = g \cos \alpha - g \sin \alpha = g(\cos \alpha - \sin \alpha)$.The length of the incline is $L = \frac{H}{\sin \alpha}$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ with initial velocity $u = 0$:$\frac{H}{\sin \alpha} = \frac{1}{2} g(\cos \alpha - \sin \alpha) t^2$.$t^2 = \frac{2H}{g \sin \alpha (\cos \alpha - \sin \alpha)} = \frac{2H}{g (\sin \alpha \cos \alpha - \sin^2 \alpha)}$.Using $2 \sin \alpha \cos \alpha = \sin 2\alpha$ and $2 \sin^2 \alpha = 1 - \cos 2\alpha$:$t^2 = \frac{4H}{g (\sin 2\alpha - (1 - \cos 2\alpha))} = \frac{4H}{g (\sin 2\alpha + \cos 2\alpha - 1)}$.For $0 Since the denominator increases, $t^2$ decreases, and thus the time $t$ decreases with $\alpha$.

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