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Q.
An object of mass m is released from rest from the top of a smooth inclined plane of height $h$. Its speed at the bottom of the plane is proportional to
Work, Energy and Power
Solution:
Let $v$ be the speed of the object at the bottom of the plane.
According to work-energy theorem
$W = \Delta K=K_{f} - K_{i}$
$mgh = \frac{1}{2}mv^{2}-\frac{1}{2}mu^{2} \,\,\left(\therefore u = 0\right)$
$mgh = \frac{1}{2}mv^{2}$ or $v = \sqrt{2gh}$
From this expression, it is clear that v is independent of the mass of an object.