Sign up to access the complete lesson and track your progress!
Unlock This CourseRemember, an unbalanced force changes the velocity of an object. In other words, it makes the object accelerate. This lesson tells you how much acceleration you get.
Two things decide the size of the acceleration:
The three quantities are linked by one equation:
force = mass × acceleration
F = m × a
You must learn this equation. It is not given on the formulae page of the exam paper.
| Quantity | Symbol | Unit |
|---|---|---|
| Unbalanced force | F | newton (N) |
| Mass | m | kilogram (kg) |
| Acceleration | a | metre per second squared (m/s2) |
Key terms:
The equation also tells you what a newton really is. A force of 1 N is the unbalanced force that gives a mass of 1 kg an acceleration of 1 m/s2.
The F in the equation is not just any force on the object. It is the unbalanced force, which is the resultant force once all the forces have been combined. If an engine pushes forwards with 500 N and friction pushes back with 200 N, the F that goes into the equation is 300 N.
The acceleration is always in the same direction as the unbalanced force. If the unbalanced force is zero, the acceleration is zero too.
For the same mass, a bigger force gives a bigger acceleration.
Push an empty wheelbarrow gently and it speeds up slowly. Push it hard and it speeds up quickly.
Acceleration is directly proportional to the unbalanced force. Double the force and you double the acceleration.
For the same force, a bigger mass gives a smaller acceleration.
Fill the wheelbarrow with bricks and give it the same push. It now speeds up much more slowly.
Acceleration is inversely proportional to the mass. Double the mass and you halve the acceleration.
You can rearrange F = m × a to find any one of the three quantities:
In every calculation write the equation, put in the numbers, then give the answer with its unit.
A golf buggy and its driver have a total mass of 600 kg. The buggy accelerates at 1.5 m/s2. Calculate the unbalanced force on the buggy.
F = m × a
F = 600 × 1.5
F = 900 N
A curling stone has a mass of 20 kg. A player pushes it across the ice so that the unbalanced force on it is 30 N. Calculate the acceleration of the stone.
a = F / m
a = 30 / 20
a = 1.5 m/s2
An unbalanced force of 300 N gives a rowing boat and its rower an acceleration of 1.2 m/s2. Calculate their total mass.
m = F / a
m = 300 / 1.2
m = 250 kg
Check your units first. The equation only works with newtons, kilograms and m/s2. A mass in grams must be divided by 1000. For example, a ball of mass 160 g has a mass of 0.16 kg. A force in kilonewtons (kN) must be multiplied by 1000, so 2.5 kN is 2500 N.
Slowing down. The equation works for deceleration too. A pickup truck of mass 2000 kg brakes with an acceleration of −4 m/s2. F = 2000 × (−4) = −8000 N. The minus sign shows that the unbalanced force acts in the opposite direction to the motion. Its size is 8000 N.
Exam questions often hide one of the numbers you need. There are two common types.
A moped and its rider have a total mass of 200 kg. The engine gives a driving force of 900 N forwards. Resistive forces of 300 N act backwards. Calculate the acceleration.
Step 1: find the unbalanced force.
F = 900 − 300 = 600 N forwards
Step 2: use a = F / m.
a = 600 / 200
a = 3 m/s2 forwards
An ice skater of mass 60 kg speeds up from 2 m/s to 8 m/s in 4 s. Calculate the unbalanced force on the skater.
Step 1: find the acceleration.
a = (v − u) / t
a = (8 − 2) / 4 = 1.5 m/s2
Step 2: use F = m × a.
F = 60 × 1.5
F = 90 N
You can be asked about this experiment even if you have never done it.
The apparatus. A trolley sits on a long, flat runway. A string is tied to the front of the trolley and passes over a pulley at the end of the runway. Small masses hang from the other end of the string. As the masses fall, the string pulls the trolley along. A piece of card on top of the trolley passes through two light gates. A data logger uses the speed at each gate and the time between them to work out the acceleration.
Changing the force. Each 100 g hanging mass pulls with a force of about 1 N. Start with spare masses stacked on the trolley. To increase the force, move one mass from the trolley to the end of the string. This keeps the total mass that is moving the same. Record the acceleration for each force.
Changing the mass. Keep the hanging masses the same, so the force stays the same. Add masses to the trolley. Each time, record the acceleration and the total mass that is moving (the trolley, the masses on it and the hanging masses).
What you find. A graph of acceleration against force is a straight line through the origin, so acceleration is directly proportional to force. When the total moving mass is doubled, the acceleration halves, so acceleration is inversely proportional to mass.
Making it accurate. Friction slows the trolley. Tilt the runway a little, until the trolley keeps a steady speed after a gentle push with no masses hanging. Repeat each reading and take a mean.
A tuk-tuk is a small three-wheeled taxi. A tuk-tuk and its driver have a total mass of 400 kg.
(a) State the equation linking unbalanced force, mass and acceleration. [1 mark]
(b) The engine gives a driving force of 1100 N forwards. The total resistive force is 300 N backwards. Calculate the acceleration of the tuk-tuk. [3 marks]
(c) Passengers get in, so the total mass is now 640 kg. The unbalanced force is the same as in part (b). Explain how the acceleration changes. [2 marks]
(d) Later, with the passengers still on board, the tuk-tuk speeds up from 5 m/s to 11 m/s in 4 s. Calculate the unbalanced force on the tuk-tuk. [3 marks]
(a) force = mass × acceleration, or F = m × a (1)
(b) Unbalanced force = 1100 − 300 = 800 N (1)
a = F / m = 800 / 400 (1)
a = 2 m/s2 (1)
(c) The acceleration is smaller. (1) The mass is larger and the force is the same, and a = F / m, so the same force gives less acceleration. (1)
(d) a = (v − u) / t = (11 − 5) / 4 = 1.5 m/s2 (1)
F = m × a = 640 × 1.5 (1)
F = 960 N (1)
Parts (b) and (d) are both two-step problems, and each step earns its own mark. Write down the 800 N and the 1.5 m/s2 even if you can do them in your head. In part (d), use the new mass of 640 kg, not the 400 kg from the start of the question.