WAEC 2013 · Paper 2 · Q17

(a)
A lorry of mass $1200\text{ kg}$ was kept in motion by a force of $120\text{ N}$ . If the lorry was initially at rest, calculate the distance covered in 6 seconds.
(b)
An object weighing $30\text{ N}$ is kept in equilibrium by two strings inclined at $30^\circ$ and $60^\circ$ to the horizontal. Find the tensions in the strings.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

From $F = ma$ : $a = \dfrac{120}{1200}$
$= 0.1\text{ m s}^{-2}$ .
Use $s = ut + \frac12at^2$ with $u = 0$ and $t = 6$ : $s = \frac12 \times 0.1 \times 36$ .
So the lorry covers $1.8\text{ m}$ .

Let $T_1$ be the tension in the string at $60^\circ$ and $T_2$ the tension in the string at $30^\circ$ .
Resolve horizontally: $T_1\cos60^\circ = T_2\cos30^\circ$ , so $\frac12T_1 = \frac{\sqrt3}{2}T_2$ and $T_1 = \sqrt3\,T_2$ .
Resolve vertically: $T_1\sin60^\circ + T_2\sin30^\circ = 30$ .
Substitute $T_1 = \sqrt3\,T_2$ : $\sqrt3\,T_2 \times \frac{\sqrt3}{2} + \frac12T_2 = 30$ , so $2T_2 = 30$ .
So $T_2 = 15\text{ N}$ and $T_1 = 15\sqrt3 \approx 25.98\text{ N}$ .
The tensions are $25.98\text{ N}$ (string at $60^\circ$ ) and $15\text{ N}$ (string at $30^\circ$ ).