9709 · 4.4

Newton's laws of motion — practice questions

Practice and worked examples for 9709 Newton's laws of motion. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A box of mass 5 kg rests on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.4. A horizontal force of 30 N is applied to the box. Find the acceleration of the box. (Take $g = 9.8$ m s⁻²).

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Draw a force diagram. Identify all forces: Weight ( $W$ ), Normal Reaction ( $R$ ), Applied Force (30 N), and Friction ( $F_r$ ).
Resolve forces vertically. The box is not accelerating vertically, so forces are in equilibrium. $R - W = 0$ . We know $W = mg = 5 \times 9.8 = 49$ N. Therefore, $R = 49$ N.
Calculate maximum friction. The maximum frictional force is $F_{r,max} = \mu R = 0.4 \times 49 = 19.6$ N.
Check for motion. The applied force (30 N) is greater than the maximum friction (19.6 N), so the box will move. The frictional force opposing the motion will be 19.6 N.
Apply Newton's Second Law horizontally. The resultant force is in the direction of motion. $F_{net} = \text{Applied Force} - \text{Friction}$ . $F_{net} = 30 - 19.6 = 10.4$ N.
Calculate acceleration. Using $F_{net} = ma$ : $10.4 = 5 \times a$ $a = \frac{10.4}{5} = 2.08$ m s⁻².

Worked example 2

Two particles, A of mass 3 kg and B of mass 2 kg, are connected by a light inextensible string. Particle A is pulled along a smooth horizontal surface by a horizontal force of 40 N. Find the acceleration of the system and the tension in the string.

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Consider the system as a whole. Treat both particles as a single object with combined mass $m_{total} = 3 + 2 = 5$ kg. The only external horizontal force is 40 N (tension is an internal force). Apply $F=ma$ to the whole system: $40 = (3+2)a$ $40 = 5a$ $a = 8$ m s⁻².
Isolate one particle to find tension. Let's consider particle B. The only horizontal force acting on B is the tension ( $T$ ) from the string, pulling it forward. Apply $F=ma$ to particle B: $T = m_B \times a$ $T = 2 \times 8$ $T = 16$ N.
(Check) Isolate particle A. The forces on A are the 40 N pulling force and the tension ( $T$ ) pulling it back. The resultant force is $40 - T$ . Apply $F=ma$ to particle A: $40 - T = m_A \times a$ $40 - T = 3 \times 8$ $40 - T = 24$ $T = 40 - 24 = 16$ N. The results match.

Answer: The acceleration is 8 m s⁻² and the tension is 16 N.

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