Question

Consider an object of mass $m$ released from rest on the top of a smooth inclined plane of height $h$ . What will the speed be proportional to when it reaches the bottom of the inclined plane?

• A. $m^{0}$
• B. $m$
• C. $m^{2}$
• D. $m^{- 1}$

Answer 1

Answer: (A)

Let v be the speed of the object at the bottom the plane.
According to work energy theorem
$w=\Delta K=K_{f}-K_{i}$
$mgh= \, \frac{1}{2}mv^{2}- \, \frac{1}{2}mu^{2} \, \left(\right.\therefore u=0\left.\right)$
$mgh= \, \frac{1}{2}mv^{2} \, or \, v=\sqrt{2 g h}$
From this expression, it clear that v is independent the mass of an object.