A particle is released from top of a smooth inclined plane of inclination θ . Let v be the speed of the particle after moving a distance s. Then the quantity which is constant is

Question

A particle is released from top of a smooth inclined plane of inclination θ . Let v be the speed of the particle after moving a distance s. Then the quantity which is constant is (A) v2-2gs sin θ (B) v2-√2 g s sin θ (C) v2+2 g s sin θ (D) v2+√2 g s sin θ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a0710110156d64a38502eec2916e453f-.png" /> v =√2gH =√2gs sin θ so, v2-2gs sin θ=0

Q. A particle is released from top of a smooth inclined plane of inclination $θ .$ Let $v$ be the speed of the particle after moving a distance s. Then the quantity which is constant is

$v^{2} - 2 g s sin θ$

$v^{2} - 2 g s sin θ$

$v^{2} + 2 g s sin θ$

$v^{2} + 2 g s sin θ$

$v = 2 g H = 2 g s sin θ$
so, $v^{2} - 2 g s sin θ = 0$

Q. A particle is released from top of a smooth inclined plane of inclination θ. Let v be the speed of the particle after moving a distance s. Then the quantity which is constant is

v2−2gssinθ

v2−2gs​sinθ

v2+2gssinθ

v2+2gs​sinθ

v=2gH​=2gssinθ​ so, v2−2gssinθ=0

Q. A particle is released from top of a smooth inclined plane of inclination $θ .$ Let $v$ be the speed of the particle after moving a distance s. Then the quantity which is constant is

$v^{2} - 2 g s sin θ$

$v^{2} - 2 g s sin θ$

$v^{2} + 2 g s sin θ$

$v^{2} + 2 g s sin θ$

$v = 2 g H = 2 g s sin θ$
so, $v^{2} - 2 g s sin θ = 0$