3.4.1.5
Newton's second law: $F = ma relates resultant force to acceleration$
Newton's Laws & Momentum — AQA A-Level Physics
Key Definition
Newton's second law — The resultant force on an object of constant mass is directly proportional to its acceleration. F = ma.
$$F = ma$$
- $F$: resultant force (N)
- $m$: mass (kg)
- $a$: accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². (m s⁻²)
- AccelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². always acts in the same direction as the resultant force.
- If the resultant force is along the direction of motion: the object speeds up or slows down.
- If the resultant force is at an angle to the direction of motion: the object changes direction.
- The resultant force is the vector sum of all forces on the object.
Worked Example
A rocket has thrust 15 MN upwards, weight 8 MN, and air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω). 500 kN. Mass = \(0.8 \times 10^{5}\) kg. Find the resultant force and accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻²..
Show Solution
1
Calculate resultant force (upward positive)
$$F = (15 \times 10^6) + (-8 \times 10^6) + (-500 \times 10^3)$$
$$F = 6.5 \times 10^6 \text{ N} = 6.5 \text{ MN upwards}$$2
Calculate acceleration
$$a = \frac{F}{m} = \frac{6.5 \times 10^6}{0.8 \times 10^5} = 81 \text{ m s}^{-2} \text{ upwards}$$
Answer
Resultant force = 6.5 MN upwards. Acceleration = 81 m s$^{-2}$ upwards.
Examiner Tips and Tricks
- Take the direction of motion as positive.
- Opposing forces (drag, weight if moving up) are negative.
- For stationary objects, choose any consistent sign convention.