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."
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.
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."
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 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 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.
\(a = \frac{1.008 \times 10^6}{3.33 \times 10^5}\) \(a = 3.027m/s/s\) \(a \approx 3.0m/s/s\) |