WAEC 2009 · Paper 2 · Q17

A body of mass 3 kg resting on a smooth surface is acted upon by forces $P(10\text{ N}, 045^\circ)$ , $Q(12\text{ N}, 180^\circ)$ , $R(15\text{ N}, 215^\circ)$ and $S(8\text{ N}, 090^\circ)$ . Calculate, correct to one decimal place, the:

(a)
resultant force acting on the body;
(b)
acceleration with which the body begins to move;
(c)
time it takes to cover the first 4 metres.

Worked solution (try it first)

(a)

Resolve each force into east ( $x$ ) and north ( $y$ ) components, using the bearing $\theta$ : $x = F\sin\theta$ , $y = F\cos\theta$ .
East: $10\sin 45^\circ + 12\sin 180^\circ + 15\sin 215^\circ + 8\sin 90^\circ = 7.071 + 0 - 8.604 + 8$
$= 6.467$ .
North: $10\cos 45^\circ + 12\cos 180^\circ + 15\cos 215^\circ + 8\cos 90^\circ = 7.071 - 12 - 12.287 + 0$
$= -17.216$ .
Magnitude: $|\mathbf R| = \sqrt{6.467^2 + 17.216^2}$
$= \sqrt{338.21}$
$= 18.4$ N (1 d.p.).

(b)

Newton's second law, $F = ma$ : $a = \dfrac{18.39}{3} = 6.1$ m s $^{-2}$ (1 d.p.).

(c)

The body starts from rest, so $s = ut + \frac12at^2$ with $u = 0$ : $4 = \frac12(6.13)t^2$ .
$t^2 = \dfrac{8}{6.13} = 1.305$ , so $t = 1.1$ s (1 d.p.).

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