Newton’s Second Law: Force, Mass and Acceleration

Push a child's empty trolley and it shoots forward easily; pile it high with shopping and the same push barely gets it moving. Press the accelerator in a small car and you are pinned back by the surge; the same engine in a heavily loaded van produces only a gentle pull-away. Behind all of this is Newton's Second Law, the equation that ties together the three central ideas of forces: the force applied, the mass of the object, and the acceleration that results. This lesson, part of Topic P2 (Forces) of OCR Gateway Science A, sets out the law $F = ma$ , rearranges it both ways with worked calculations, and walks through the required practical that investigates how acceleration depends on force and on mass.

By the end of this lesson you should be able to state Newton's Second Law, use and rearrange the equation $F = ma$ , describe the required practical investigating how acceleration depends on force and on mass, identify its variables and errors, and relate the law to inertial mass.

Newton’s Second Law

Newton's Second Law states that the acceleration of an object is proportional to the resultant force acting on it, and inversely proportional to its mass. In words: a bigger force gives a bigger acceleration, but a bigger mass gives a smaller acceleration for the same force. The law is written as the equation:

$F = ma$

where $F$ is the resultant force (in $\text{N}$ ), $m$ is the mass (in $\text{kg}$ ) and $a$ is the acceleration (in $\text{m/s}^2$ ).

Two points are worth stressing. First, $F$ is the resultant (overall) force — if several forces act, you must work out the resultant first. Second, the acceleration is in the same direction as the resultant force: push something to the right and it accelerates to the right.

The equation captures both parts of the law at once:

for a fixed mass, doubling the force doubles the acceleration (acceleration ∝ force);
for a fixed force, doubling the mass halves the acceleration (acceleration ∝ 1/mass).

Exam Tip: $F = ma$ uses the resultant force in newtons, mass in kilograms, and acceleration in $\text{m/s}^2$ . The acceleration is always in the same direction as the resultant force. Convert any mass given in grams to kg before substituting.

Rearranging $F = ma$

The equation can be rearranged to make the mass or the acceleration the subject:

$F = ma \qquad a = \dfrac{F}{m} \qquad m = \dfrac{F}{a}$

A formula triangle is a useful aid: cover the quantity you want, and the triangle shows the calculation.

Force sits at the top, so $F = ma$ ; mass and acceleration sit side by side at the bottom, so $a = \dfrac{F}{m}$ and $m = \dfrac{F}{a}$ .

Worked Example 1 — Finding the force

A car of mass $1200 \text{ kg}$ accelerates at $2 \text{ m/s}^2$ . Calculate the resultant force needed.

Step 1 — write the equation: $F = ma$ .

Step 2 — substitute: $F = 1200 \times 2$ .

Step 3 — calculate: $F = 2400 \text{ N}$ .

Answer: the resultant force is $2400 \text{ N}$ .

Worked Example 2 — Finding the acceleration

A resultant force of $6000 \text{ N}$ acts on a van of mass $1500 \text{ kg}$ . Calculate its acceleration.

Step 1 — rearrange for acceleration: $a = \dfrac{F}{m}$ .

Step 2 — substitute: $a = \dfrac{6000}{1500}$ .

Step 3 — calculate: $a = 4 \text{ m/s}^2$ .

Answer: the acceleration is $4 \text{ m/s}^2$ .

Worked Example 3 — Finding the mass

A resultant force of $200 \text{ N}$ gives a trolley an acceleration of $2.5 \text{ m/s}^2$ . Calculate the mass of the trolley.

Step 1 — rearrange for mass: $m = \dfrac{F}{a}$ .

Step 2 — substitute: $m = \dfrac{200}{2.5}$ .

Step 3 — calculate: $m = 80 \text{ kg}$ .

Answer: the mass of the trolley is $80 \text{ kg}$ .

Worked Example 4 — Using the resultant force

A box of mass $5 \text{ kg}$ is pushed with a force of $30 \text{ N}$ while friction of $10 \text{ N}$ opposes the motion. Calculate the acceleration.

Step 1 — find the resultant force first: $F = 30 - 10 = 20 \text{ N}$ .

Step 2 — rearrange for acceleration: $a = \dfrac{F}{m}$ .

Step 3 — substitute and calculate: $a = \dfrac{20}{5} = 4 \text{ m/s}^2$ .

Answer: the acceleration is $4 \text{ m/s}^2$ . Note that you must use the resultant force ( $20 \text{ N}$ ), not the applied force ( $30 \text{ N}$ ).

Exam Tip: When other forces (like friction) act, find the resultant force first, then put it into $F = ma$ . Using the applied force instead of the resultant is one of the commonest mistakes in these questions.

Required Practical: How Acceleration Depends on Force and on Mass

A key P2 required practical is to investigate how the acceleration of a trolley depends on (a) the force applied to it and (b) the mass being accelerated. The trolley is pulled along a runway and its acceleration is measured, while either the force or the mass is changed.

The acceleration can be measured using light gates connected to a data logger (which times the trolley through known distances and computes the acceleration) or, in a more traditional set-up, ticker tape through a ticker-timer (where the spacing of the dots reveals how the speed changes).

Method (numbered):

Newton’s Second Law: Force, Mass and Acceleration

Newton’s Second Law: Force, Mass and Acceleration