A-level · Forces & Newton's laws

Newton's laws and F = maExam-style questions

Newton's three laws and the one equation that runs all of dynamics — the RESULTANT force equals mass times acceleration — applied along a line and to forces given as 2-D vectors, why weight is mg and the normal reaction is whatever balances the rest, and how a lift problem is just F = ma done vertically.

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12 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
A force of $6$ N acts on a particle of mass $2$ kg. Find the acceleration.
Level 2 · 2 marksTry it
2
A particle of mass $5$ kg accelerates at $4\ \text{m s}^{-2}$ . Find the resultant force on it.
Level 2 · 2 marksTry it
3
Figure 1Diagram not accurately drawn.

A lift of mass $M$ hangs from a vertical cable which is fixed to the centre of its roof. The cable is used to raise and lower the lift.

Two crates are stacked inside the lift. Crate $A$ , of mass $3m$ , stands on the floor of the lift and crate $B$ , of mass $2m$ , stands on top of crate $A$ , as shown in Figure 1.

In a model of the motion, the cable is light and inextensible, there is no air resistance and each crate is a particle.

The lift is moving vertically downwards and is slowing down with deceleration of magnitude $\frac{g}{4}$ . The tension in the cable is $20mg$ .

Using the model,
(a) find $M$ in terms of $m$ ,

(4 marks)
(b) find, in terms of $m$ and $g$ , the magnitude of the force exerted on crate $A$ by crate $B$ .

(3 marks)
Level 3 · 7 marksTry it
4
A woman of mass $60$ kg stands on the floor of a lift. The lift has mass $400$ kg and is raised by a vertical cable.

Take $g = 9.8$ m s⁻².
(a) Find the tension in the cable when the lift is accelerating upwards at $1.2$ m s⁻². (3 marks)
(b) Find the magnitude of the force exerted by the floor of the lift on the woman at that instant. (3 marks)
(c) Later, while still moving upwards, the lift decelerates at $1.5$ m s⁻². Find the magnitude of the force exerted by the floor on the woman now. (2 marks)
Level 3 · 8 marksTry it
5
A particle is acted on by the forces

$\mathbf{P} = (-3\mathbf{i} + 8\mathbf{j})\ \text{N} \qquad \mathbf{Q} = (2k\mathbf{i} - 3k\mathbf{j})\ \text{N}$

where $k$ is a constant. No other forces act on the particle.

The particle moves with acceleration $(4\mathbf{i} + \mathbf{j})\ \text{m s}^{-2}$

Find $k$

(4 marks)
Level 3 · 4 marksTry it

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