A body of mass m = 1 kg , at rest on a smooth horizontal…

A body of mass $m = 1\ \mathrm{kg}$, at rest on a smooth horizontal surface, is acted on by a force of constant magnitude $F = mg/2$. As the body moves in a straight line, the angle $\theta$ between the force and the horizontal changes as $\theta = \pi s/k$, where $s$ is the distance travelled and $k = 8\ \mathrm{m}$. What is the body's speed when $\theta = \pi/2$?

Explanation of Method

A force that turns as the body moves.

Step 1: Only the horizontal part accelerates the body

The vertical part $F\sin\theta \le mg/2$ is always less than the weight, so the body stays on the surface. Along the surface: $m\,v\frac{dv}{ds} = F\cos\theta = \frac{mg}{2}\cos\frac{\pi s}{k}$.

Step 2: Integrate over distance

\[ \frac{v^2}{2} = \frac{g}{2}\int_0^s \cos\frac{\pi s'}{k}\,ds' = \frac{gk}{2\pi}\sin\frac{\pi s}{k}, \]

\[ v^2 = \frac{gk}{\pi}\sin\theta. \]

Step 3: At $\theta = \pi/2$

This happens at $s = k/2 = 4\ \mathrm{m}$, where $\sin\theta = 1$: $u = \sqrt{\frac{gk}{\pi}} = \sqrt{\frac{9.81 \cdot 8}{\pi}}$.

\[u \approx 5.0\ \mathrm{m/s}\]

$\theta = 0$ $30^\circ$ $60^\circ$ $90^\circ$ $s$ $\theta = \frac{\pi s}{k},\ F = \frac{mg}{2}$ $v^2 = \frac{gk}{\pi}\sin\theta$ $v = 5.0\ \mathrm{m/s}$

The force keeps its size but turns upward as the body moves; only its horizontal part $F\cos\theta$ speeds the body up. At $\theta = 90^\circ$ (after $s = k/2 = 4\ \mathrm{m}$) it no longer pushes forward at all.

Result

\[\boxed{\,u \approx 5.0\ \mathrm{m/s}\,}\]

Remark

About the original solution. It read the law as $\theta = \pi k s$ with $k = 8\ \mathrm{m}$, which does not work dimensionally ($\theta$ would be in square metres), and got $u \approx 0.62\ \mathrm{m/s}$ at $s = 1/16\ \mathrm{m}$. With $k$ given in metres, the angle must be $\theta = \pi s/k$, which gives 5.0 m/s. (If $k$ were meant as $8\ \mathrm{m^{-1}}$, the formula $u = \sqrt{g/(\pi k)} \approx 0.62\ \mathrm{m/s}$ would apply.) The method, $v\,dv/ds = a$, is the same in both cases.