An object of mass m is allowed to fall from rest along a rough inclined plane. The speed of the object reaching the bottom of the plane is proportional to

Question

An object of mass m is allowed to fall from rest along a rough inclined plane. The speed of the object reaching the bottom of the plane is proportional to (A) m  (B) m2 (C) m-1 (D) m0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bdc413fa53b40b419fbefd1a690c6edf-.png" /> From work energy principle, (1/2) mv2 = mgh - f × s =mgh - (μ mg cos θ ) × (h/ sin θ) v = √2gh - 2 μ gh cot θ Hence, v propto m0

Q. An object of mass $m$ is allowed to fall from rest along a rough inclined plane. The speed of the object reaching the bottom of the plane is proportional to

$m$

$m^{2}$

$m^{- 1}$

$m^{0}$

From work energy principle,
$\frac{1}{2} m v^{2} = m g h - f \times s$
$= m g h - (μ m g cos θ) \times \frac{h}{s i n θ}$
$v = 2 g h - 2 μg h cot θ$
Hence, $v \propto m^{0}$

Q. An object of mass m is allowed to fall from rest along a rough inclined plane. The speed of the object reaching the bottom of the plane is proportional to

m

m2

m−1

m0

From work energy principle, 21​mv2=mgh−f×s =mgh−(μmgcosθ)×sinθh​ v=2gh−2μghcotθ​ Hence, v∝m0

Q. An object of mass $m$ is allowed to fall from rest along a rough inclined plane. The speed of the object reaching the bottom of the plane is proportional to

$m$

$m^{2}$

$m^{- 1}$

$m^{0}$

From work energy principle,
$\frac{1}{2} m v^{2} = m g h - f \times s$
$= m g h - (μ m g cos θ) \times \frac{h}{s i n θ}$
$v = 2 g h - 2 μg h cot θ$
Hence, $v \propto m^{0}$