A small body of mass m lies on a smooth horizontal surface.…

A small body of mass $m$ lies on a smooth horizontal surface. From $t = 0$ a force $\vec F$ acts on it; its magnitude grows as $F = kt$ ($k$ constant) and its direction is fixed, at an angle $\theta$ above the horizontal. (a) What is the body's speed when it leaves the surface? (b) How far has it travelled by then?

Explanation of Method

When a body pulled by a growing force lifts off.

Step 1: When it leaves the surface

While it is on the surface, the vertical forces balance: $N + kt\sin\theta = mg$. The body lifts off when $N$ drops to zero: $kt_0\sin\theta = mg \;\Longrightarrow\; t_0 = \frac{mg}{k\sin\theta}$.

Step 2: (a) Speed at that moment

Until then the body moves horizontally only: $m\frac{dv}{dt} = kt\cos\theta \;\Longrightarrow\; v = \frac{k\cos\theta}{2m}\,t^2$.

\[v_0 = \frac{k\cos\theta}{2m}\,t_0^2 = \frac{mg^2\cos\theta}{2k\sin^2\theta}\]

Step 3: (b) Distance

\[ s = \int_0^{t_0} v\,dt = \frac{k\cos\theta}{6m}\,t_0^3. \]

\[s = \frac{m^2g^3\cos\theta}{6k^2\sin^3\theta}\]

$F = kt$ $mg$ $N$ $\theta$ $N$ $t$ $t_0$ $mg$ $N = mg - kt\sin\theta$

As the force grows, its upward part $kt\sin\theta$ takes over more of the weight, and the floor's push $N$ falls. It reaches zero at $t_0 = mg/(k\sin\theta)$: the body lifts off.

Result

\[\boxed{\,v_0 = \frac{k\cos\theta}{2m}\,t_0^2 = \frac{mg^2\cos\theta}{2k\sin^2\theta}\,}\]

\[\boxed{\,s = \frac{m^2g^3\cos\theta}{6k^2\sin^3\theta}\,}\]

Remark

About the original solution. It gave the body a vertical acceleration $kt\sin\theta/m$ from the start, ignoring gravity and the surface, and then looked for the moment when the vertical velocity is zero, which led nowhere ($t = 0$); no final answer was obtained. The key is that the body stays on the surface until the upward part of the force equals its weight; after that it lifts off.