# Newton's Second Law

## Key Stage 4 Foundation

### Meaning

Newton's Second Law states that "Force = Mass x Acceleration, which means the acceleration of an object is directly proportional to the resultant force acting upon it."

Newton's Second Law can be used to calculate the acceleration of an object given its mass and the resultant force acting upon it.

### Equation

Force = Mass x Acceleration

$$F=ma$$

Where

$$F$$ = The Resultant Force on the object.

$$m$$ = The mass of the object.

$$a$$ = The acceleration of the object.

### Example Calculations

#### Finding the Force given Mass and Acceleration

 A 2.3kg object accelerates at a rate of 8.8m/s/s. Calculate the resultant force acting on the object correct to two significant figures. A 5.5x104kg rocket accelerates at a rate of 61m/s/s. Calculate the resultant force acting on the rocket correct to two significant figures. 1. State the known quantities m = 2.3kg a = 8.8m/s/s 1. State the known quantities m = 5.5x104kg a = 61m/s/s 2. Substitute the numbers into the equation and solve. $$F=ma$$ $$F=2.3 \times 8.8$$ $$F=20.24N$$ $$F \approx 20N$$ 2. Substitute the numbers into the equation and solve. $$F=ma$$ $$F=5.5 \times 10^4 \times 61$$ $$F=3355000N$$ $$F \approx 3.4 \times 10^6 N$$

#### Finding the Acceleration given Mass and Force

 A 7kg object is subjected to a resultant force of 53N. Calculate the acceleration of the object correct to two significant figures. A 160g snooker ball experiences a resultant force of 12N. Calculate the acceleration of the snooker ball correct to two significant figures. 1. State the known quantities m = 7kg F = 53N 1. State the known quantities m = 160g = 0.16kg F = 12N 2. Substitute the numbers and evaluate. $$F=ma$$ $$53=7a$$ 2. Substitute the numbers and evaluate. $$F=ma$$ $$12=0.16a$$ 3. Rearrange the equation and solve. $$a = \frac{53}{7}$$ $$a = 7.571m/s/s$$ $$a \approx 7.6m/s/s$$ 3. Rearrange the equation and solve. $$a = \frac{12}{0.16}$$ $$a = 75m/s/s$$

## Key Stage 4 Higher

### Meaning

Newton's Second Law states that "Force = Mass x Acceleration, which means the acceleration of an object is directly proportional to the resultant force acting upon it."

Newton's Second Law can be used to calculate the acceleration of an object given its mass and the resultant force acting upon it.
Newton's Second Law provides a definition for inertial mass as the ratio of force to the acceleration of an object $$m= \frac{F}{a}$$.

### Equation

Force = (Inertial Mass) x Acceleration

$$F=ma$$

Where

$$F$$ = The Resultant Force on the object.

$$m$$ = The Inertial Mass of the object.

$$a$$ = The acceleration of the object.

### Example Calculations

#### Finding the Force given Inertial Mass and Acceleration

 A 1.23Mg object accelerates at a rate of 5.3x10-2m/s/s. Calculate the resultant force acting on the object correct to two significant figures. A 1.7x10ng cell accelerates at a rate of 320m/s/s. Calculate the resultant force acting on the rocket correct to two significant figures. 1. State the known quantities m = 1.23Mg = 1230kg a = 5.3x10-2m/s/s 1. State the known quantities m = 1.7ng = 1.7x10-12kg a = 320m/s/s. 2. Substitute the numbers into the equation and solve. $$F=ma$$ $$F=1230 \times 5.3 \times 10^{-2}$$ $$F=65.19$$ $$F \approx 65N$$ 2. Substitute the numbers into the equation and solve. $$F=ma$$ $$F=1.7 \times 10^{-12} \times 320$$ $$F=0.000000000544N$$ $$F \approx 5.4 \times 10^{-10}N$$

#### Finding the Acceleration given Inertial Mass and Force

 A 9.2g object is subjected to a resultant force of 3.7kN. Calculate the acceleration of the object correct to two significant figures. A 333 tonne passenger plane experiences a resultant force of 1.008MN. Calculate the acceleration of the passenger plane correct to two significant figures. 1. State the known quantities m = 9.2g = 0.0092kg F = 3.7kN = 3700N 1. State the known quantities m = 333 tonne = 3.33 x 105kg F = 1.008MN = 1.008 x 106N 2. Substitute the numbers and evaluate. $$F=ma$$ $$3700=0.0092a$$ 2. Substitute the numbers and evaluate. $$F=ma$$ $$1.008 \times 10^6 = (3.33 \times 10^5 )a$$ 3. Rearrange the equation and solve. $$a = \frac{3700}{0.0092}$$ $$a = 402173.913m/s/s$$ $$a \approx 4.0 \times 10^5m/s/s$$ 3. Rearrange the equation and solve. $$a = \frac{1.008 \times 10^6}{3.33 \times 10^5}$$ $$a = 3.027m/s/s$$ $$a \approx 3.0m/s/s$$

#### Finding the Inertial Mass given the Force and Acceleration

 An object is subjected to a resultant force of 92N and accelerates at a rate of 0.42m/s/s. Calculate the inertial mass of the object correct to two significant figures. The brakes of a car provide a force of 12kN and are able to decelerate it at a rate of 8.7m/s/s. Calculate the intertial mass of the car correct to two significant figures. 1. State the known quantities a = 0.42m/s/s F = 92N 1. State the known quantities a = 8.7m/s/s F = 12kN = 12 x 103N 2. Substitute the numbers and evaluate. $$F=ma$$ $$92=0.42m$$ 2. Substitute the numbers and evaluate. $$F=ma$$ $$12 \times 10^3 = 8.7m$$ 3. Rearrange the equation and solve. $$m = \frac{92}{0.42}$$ $$m = 219.047619kg$$ $$m \approx 220kg$$ 3. Rearrange the equation and solve. $$m = \frac{12 \times 10^3}{8.7}$$ $$m = 1379.31034kg$$ $$m \approx 1400kg$$

### References

#### AQA

Newton’s Second Law of motion, pages 144-145, GCSE Physics; Third Edition, Oxford University Press, AQA
Newton’s Second Law, pages 164, 165, GCSE Combined Science Trilogy; Physics, CGP, AQA
Newton’s Second Law, pages 196, 197, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Newton’s Second Law, pages 212, 214, GCSE Combined Science; The Revision Guide, CGP, AQA
Newton’s Second Law; investigating, page 214, GCSE Combined Science; The Revision Guide, CGP, AQA
Newton’s Second Law; investigating, pages 167, 168, GCSE Combined Science Trilogy; Physics, CGP, AQA
Newton’s Second Law; investigating, pages 199, 200, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA

#### Edexcel

Newton’s Second Law, pages 149, 151, 154, GCSE Combined Science; The Revision Guide, CGP, Edexcel
Newton’s Second Law, pages 16, 18, 21, GCSE Physics; The Revision Guide, CGP, Edexcel
Newton’s Second Law, pages 18-19, GCSE Physics, Pearson Edexcel
Newton’s Second Law, pages 302-303, GCSE Combined Science, Pearson Edexcel
Newton’s Second Law, pages 35, 36, 46, GCSE Physics, CGP, Edexcel
Newton’s Second Law; investigating, pages 39, 40, GCSE Physics, CGP, Edexcel