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?

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) m0 (B) m (C) m2 (D) m- 1

Tardigrade Admin · Accepted Answer

Let v be the speed of the object at the bottom the plane.
According to work energy theorem

w = Δ K = K_{f} - K_{i}

m g h = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} (∴ u = 0)

m g h = \frac{1}{2} m v^{2} or v = 2 g h

From this expression, it clear that v is independent the mass of an object.

Q. 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?

$m^{0}$

$m$

$m^{2}$

$m^{- 1}$

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

Q. 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?

m0

m

m2

m−1

Let v be the speed of the object at the bottom the plane. According to work energy theorem w=ΔK=Kf​−Ki​ mgh=21​mv2−21​mu2(∴u=0) mgh=21​mv2orv=2gh​ From this expression, it clear that v is independent the mass of an object.

Q. 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?

$m^{0}$

$m$

$m^{2}$

$m^{- 1}$