Newton's Second Law
Contents
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."
About Newton's Second Law
- 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 5.5x10^{4}kg 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 = 5.5x10^{4}kg 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."
About Newton's Second Law
- 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^{-2}m/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^{-2}m/s/s |
1. State the known quantities
m = 1.7ng = 1.7x10^{-12}kg 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 10^{5}kg F = 1.008MN = 1.008 x 10^{6}N |
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 10^{3}N |
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