Newton's second law
Force, mass and acceleration
Newton's second law of motion can be described by this equation:
resultant force = mass × acceleration
\(\text{F} = \text{ma}\)
This is when:
- Force (\(\text{F}\)) is measured in newtons (N)
- Mass (\(\text{m}\)) is measured in kilograms (kg)
- Acceleration (\(\text{a}\)) is measured in metres per second squared (m/s²)
The equation shows that the acceleration of an object is:
- proportionalA relationship between two variables, eg in a gas. As temperature increases, the pressure would also increase proportionally. (If the temperature doubled, the pressure would double). to the resultant forceThe single force that could replace all the forces acting on an object, found by adding these together. If all the forces are balanced, the resultant force is zero. on the object. The symbol α is often used to represent proportionality. So, to a scientist, \(\text{a α F}\), means acceleration is proportional to the force.
- inversely proportionalA relationship between two variables where as one variable increases, the other variable decreases, eg as the volume doubled, the pressure decreased by half. to the mass of the object (\(\text{a} \alpha \frac{1}{\text{m}}\)).
In other words, the acceleration of an object increases if the resultant force on it increases, and decreases if the mass of the object increases.
Example
Calculate the force needed to accelerate a 22 kg cheetah at 15 m/s².
\(\text{F} = \text{ma}\)
= 22 × 15
= 330 N
Question
Calculate the force needed to accelerate a 15 kg gazelle at 10 m/s².
\(\text{F} = \text{ma}\)
= 15 × 10
= 150 N
Estimations
It is important to be able to estimate speeds, accelerations and forces involved in road vehicles. The symbol ~ is used to indicate that a value or answer is an approximate one. The table gives some examples.
| Vehicle | Maximum legal speed on a single carriageway in m/s | Mass in kg | Acceleration in m/s2 |
| Family car | ~27 | ~1,600 | ~3 |
| Lorry | ~22 | ~36,000 | ~0.4 |
| Vehicle | Family car |
|---|---|
| Maximum legal speed on a single carriageway in m/s | ~27 |
| Mass in kg | ~1,600 |
| Acceleration in m/s2 | ~3 |
| Vehicle | Lorry |
|---|---|
| Maximum legal speed on a single carriageway in m/s | ~22 |
| Mass in kg | ~36,000 |
| Acceleration in m/s2 | ~0.4 |
Example
Estimate the force needed to accelerate a family car to its top speed on a single carriageway.
Using values of ~1,600 kg and ~3 m/s2, and \(\text{F} = \text{ma}\):
1,600 × 3 = ~4,800 N
Question
Estimate the force needed to accelerate a lorry to its top speed on a single carriageway.
Using values of ~36,000 kg and ~0.4 m/s2, and \(\text{F} = \text{ma}\):
Force (\(\text{F}\)) is ~14,400 N
Acceleration when considering more than one force
Acceleration occurs if there is a resultant force. If there is only one force, then that is the resultant force. But if there are two or more forces acting, it is important to realise that of all the acting forces are resultant, which causes a body to accelerate.
Example
A 2 kg box of biscuits is pushed across a table with a force of 10 N. There is a frictional force of 4 N. What will be the acceleration?
First find the resultant force: This will be 6 N because the 10 N and 4 N forces oppose each other.
Next use \(\text{F} = \text{ma}\) but rearrange this to find acceleration.
\(\text{F} = \text{ma}\) becomes \(\text{a} = \frac{\text{F}}{\text{m}}\)
Acceleration = 6 ÷ 2 = 3 m/s2
Question
A 200 kg motorbike experiences a thrust of 100 N and total resistive forces of 50 N. What would be the acceleration?
Resultant force = 100 – 50 = 50 N
\(\text{a} = \frac{\text{F}}{\text{m}}\)
acceleration = 50 ÷ 200 = 0.25 m/s2
More guides on this topic
- Motion - AQA Synergy
- Circuits - AQA Synergy
- Mains electricity - AQA Synergy
- Acids and alkalis - AQA Synergy
- Rates of reaction - AQA Synergy
- Energy, rates and reactions - AQA Synergy
- Equilibria - AQA Synergy
- Electrons and chemical reactions - AQA Synergy
- Sample exam questions - movement and interactions - AQA Synergy
