3.2 Forces, acceleration and Newton's laws of motion — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

A bus speeds up from 10 m/s to 16 m/s in 3.0 s. Calculate its acceleration.
A force of 30 N stretches a spring by 0.15 m. Calculate the spring constant.

Answers

$a = \dfrac{\Delta v}{t} = \dfrac{16 - 10}{3.0} = \dfrac{6.0}{3.0} = 2.0\,\text{m/s}^2$ .
$k = \dfrac{F}{e} = \dfrac{30}{0.15} = 200\,\text{N/m}$ .

If question 1 went wrong, remember that acceleration uses the change in velocity. If question 2 went wrong, look again at Section 2.2.

Practise with help

1. A car of mass 1500 kg accelerates at $3.0\,\text{m/s}^2$ . Calculate the resultant force.

Hint

Newton's second law links the resultant force, the mass and the acceleration.

Next step

Write $F = ma$ , then multiply 1500 kg by $3.0\,\text{m/s}^2$ . The mass is already in kilograms.

Full working

\begin{aligned} F &= m a \\ &= 1500 \times 3.0 \\ &= 4500\,\text{N} \\ &= 4.5 \times 10^3\,\text{N} \end{aligned}

2. A resultant force of 90 N gives an object an acceleration of $6.0\,\text{m/s}^2$ . Calculate the mass of the object.

Hint

Rearrange $F = ma$ to make the mass the subject.

Next step

Divide the force by the acceleration: $m = \dfrac{F}{a}$ .

Full working

\begin{aligned} m &= \frac{F}{a} \\ &= \frac{90}{6.0} \\ &= 15\,\text{kg} \end{aligned}

3. A resultant force of 200 N acts on a trolley of mass 50 kg. Calculate the acceleration.

Hint

The same equation gives the acceleration when you divide the force by the mass.

Next step

Write $a = \dfrac{F}{m}$ , then divide 200 N by 50 kg.

Full working

\begin{aligned} a &= \frac{F}{m} \\ &= \frac{200}{50} \\ &= 4.0\,\text{m/s}^2 \end{aligned}

4. A car travels along a straight road at a steady 15 m/s. Explain why the resultant force on the car is zero, and what would happen if the driving force were larger than the resistive forces.

Hint

Use Newton's first law. Think about the driving force and the resistive forces.

Next step

Say what the resultant force is when the two forces are equal, then say what a non-zero resultant force does to the velocity.

Full working

The engine gives a driving force forwards, and friction and air resistance give resistive forces backwards. At a steady speed the two are equal and opposite, so the resultant force is zero. Newton's first law says that an object with a zero resultant force keeps a constant velocity. If the driving force were larger, the resultant force would be forwards, so the car would accelerate. This answer is a model answer for you to compare with your own wording.

5. A swimmer pushes backwards against the water. Use Newton's third law to explain why the swimmer moves forwards.

Hint

Newton's third law is about two objects interacting. Name both objects.

Next step

Say which force the water exerts on the swimmer, and compare its size with the force from the swimmer.

Full working

The swimmer pushes backwards on the water, and the water pushes forwards on the swimmer. By Newton's third law the two forces are equal in size and opposite in direction, and they act on different objects. The forwards force from the water is the force that moves the swimmer forwards. This answer is a model answer for you to compare with your own wording.

6. Describe how to investigate the effect of the force on the acceleration of a trolley of constant mass.

Hint

Name the equipment that measures the velocity, and state what you keep the same and what you change.

Next step

Say how you obtain the acceleration from the measurements, and say how you make the force larger.

Full working

Set up a trolley on a runway with a string over a pulley to a hanging mass. The weight of the hanging mass is the force on the trolley. Measure the velocity with light gates or a motion sensor, and find the acceleration from a velocity-time graph. Keep the mass of the trolley constant, increase the hanging mass to increase the force, and record the acceleration each time. Plot a graph of acceleration against force. This answer is a model answer for you to compare with your own wording.