At the instant t=0, a force F=kt ( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface. The time, when the body leaves the surface is

Question

At the instant t=0, a force F=kt ( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface. The time, when the body leaves the surface is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-e5enpd87gp9n.png" /> (A) mgksin α  (B) (k sin α /m g) (C) (m g sin α /k) (D) (m g/k sin α )

Tardigrade Admin · Accepted Answer

N+F sin α =mg The body leaves the surface when the normal reaction on it becomes zero. N=0 ⇒ kt sin α =mg t=(m g/k sin α )

Q. At the instant $t = 0,$ a force $F = k t$ ( $k$ is a constant) acts on a small body of mass $m$ resting on a smooth horizontal surface. The time, when the body leaves the surface is

$m g k s in α$

$\frac{k s in α}{m g}$

$\frac{m g s in α}{k}$

$\frac{m g}{k s in α}$

$N + F s in α = m g$
The body leaves the surface when the normal reaction on it becomes zero.
$N = 0$
$\Rightarrow k t s in α = m g$
$t = \frac{m g}{k s in α}$

Q. At the instant t=0, a force F=kt ( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface. The time, when the body leaves the surface is

mgksinα

mgksinα​

kmgsinα​

ksinαmg​