Newton's Laws of Motion

Lessons

Senior 1 · Physics

Newton's Laws of Motion

5 lessons

Lesson 2 of 5

Newton's Second Law: F = ma

25 min

Same engine, different load

A pick-up truck leaves a market in Muhanga empty, and pulls away from the traffic lights quickly. On the way back it carries two tonnes of bananas. The driver presses the accelerator just as hard, but now the truck gathers speed slowly. At a football match, a gentle pass and a hard shot start from the same ball at rest, but the hard shot leaves the foot far faster.

So the change in motion depends on two things: how big the resultant force is and how much mass is being moved. Newton's second law says exactly how.

Definition

Acceleration ( $a$ ) is the rate of change of velocity (Unit 2): $a = \frac{v - u}{t}$ . SI unit: metres per second squared, $\mathrm{m\,s^{-2}}$ .

Newton's second law of motion: the acceleration of a body is directly proportional to the resultant force on it, is inversely proportional to its mass, and is in the direction of the resultant force.

F = m a

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

Experiment: how acceleration depends on force and on mass

You need: a runway (a long smooth plank), a trolley, identical elastic bands, a ticker timer with paper tape (or a stopwatch and a metre rule), and extra masses to load the trolley.

Raise one end of the runway with a block until the trolley, given a small push, rolls down at a steady speed. Friction is now cancelled, so the band's pull is the resultant force.
Hook one elastic band on the trolley and pull it down the runway, keeping the band stretched to the same length all the way. Measure the acceleration from the tape.
Repeat with two and then three bands side by side, each stretched to the same length. Two bands give twice the force, three give three times.
Keep the force the same (one band) and double, then triple, the mass by stacking identical trolleys.

Figure 2.1: The block cancels friction; the stretched band gives a constant pull; the ticker timer prints a dot every fiftieth of a second on the tape, so the spacing of the dots shows how the speed changes.

Important

Safety. Put a buffer or a box of sand at the end of the runway so the trolley cannot fall off the bench onto someone's feet. Do not over-stretch elastic bands near your eyes.

Sample results

Mass kept at 1.0 kg
Resultant force (N)	Acceleration ( $\mathrm{m\,s^{-2}}$ )	$\frac{F}{a}$ (kg)
0.5	0.49	1.0
1.0	1.02	1.0
1.5	1.48	1.0
2.0	2.01	1.0

Force kept at 1.0 N
Mass (kg)	Acceleration ( $\mathrm{m\,s^{-2}}$ )	$m \times a$ (N)
0.5	2.0	1.0
1.0	1.0	1.0
2.0	0.50	1.0
3.0	0.33	1.0

Figure 2.2: Acceleration against resultant force. Each mass gives a straight line through the origin, so a is proportional to F. Doubling the mass halves the slope.

From the results to the law

The first table shows that doubling the force doubles the acceleration: $a \propto F$ . The second shows that doubling the mass halves the acceleration: $a \propto \frac{1}{m}$ . Together,

a \propto \frac{F}{m} \quad\Rightarrow\quad F = kma

The newton is defined so that the constant $k$ is exactly 1. That gives $F = m a$ .

Definition

One newton (1 N) is the resultant force that gives a mass of 1 kg an acceleration of $1\,\mathrm{m\,s^{-2}}$ .

1\,\mathrm{N} = 1\,\mathrm{kg} \times 1\,\mathrm{m\,s^{-2}} = 1\,\mathrm{kg\,m\,s^{-2}}

Worked example 1: the force to accelerate a car

A car of mass $1200\,\mathrm{kg}$ accelerates at $2.5\,\mathrm{m\,s^{-2}}$ . What resultant force acts on it?

F = ma = 1200\,\mathrm{kg} \times 2.5\,\mathrm{m\,s^{-2}} = 3000\,\mathrm{N}

Worked example 2: kicking a football

A football of mass $450\,\mathrm{g}$ is kicked with an average force of $90\,\mathrm{N}$ . Find its acceleration while the boot is touching it.

First convert the mass to kilograms: $m = 450\,\mathrm{g} = 0.45\,\mathrm{kg}$ .

a = \frac{F}{m} = \frac{90\,\mathrm{N}}{0.45\,\mathrm{kg}} = 200\,\mathrm{m\,s^{-2}}

If the boot touches the ball for $0.010\,\mathrm{s}$ , the ball leaves at $v = u + at = 0 + 200 \times 0.010 = 2.0\,\mathrm{m\,s^{-1}}$ : a gentle pass. A hard shot needs a force of well over $1000\,\mathrm{N}$ .

Worked example 3: using the resultant force

A cyclist and her bicycle have a total mass of $60\,\mathrm{kg}$ . She pushes the pedals to give a forward force of $200\,\mathrm{N}$ ; friction and air resistance total $80\,\mathrm{N}$ . Find her acceleration.

F_{\text{resultant}} = 200\,\mathrm{N} - 80\,\mathrm{N} = 120\,\mathrm{N}

a = \frac{F}{m} = \frac{120\,\mathrm{N}}{60\,\mathrm{kg}} = 2.0\,\mathrm{m\,s^{-2}}

Warning

Two common mistakes with $F = m a$

$F$ is the resultant force. In Worked example 3, using $200\,\mathrm{N}$ instead of $120\,\mathrm{N}$ gives $3.3\,\mathrm{m\,s^{-2}}$ , which is wrong.
Mass must be in kilograms. $450\,\mathrm{g}$ is $0.45\,\mathrm{kg}$ , not 450.

The second law on the road

Braking is a resultant force backwards, and the same law applies. Suppose a lorry must slow down at $3.0\,\mathrm{m\,s^{-2}}$ to stop safely before a bend on the Kigali–Rubavu road.

Lorry	Mass (kg)	Braking force needed $F = m a$
empty	10 000	$10\,000 \times 3.0 = 30\,000\,\mathrm{N}$
loaded	20 000	$20\,000 \times 3.0 = 60\,000\,\mathrm{N}$
overloaded	26 000	$26\,000 \times 3.0 = 78\,000\,\mathrm{N}$

The brakes can only supply a certain maximum force. If the load is too great, the brakes cannot give the deceleration needed, the lorry takes longer to stop, and on a long downhill it may not stop at all. This is one reason why roads have weight limits and why the police weigh lorries. The same law explains why racing cyclists ride very light bicycles: a smaller mass gives a bigger acceleration for the same push on the pedals.

Tip

Rearranging: $F = m a$ , $a = \frac{F}{m}$ , $m = \frac{F}{a}$ . Write the formula, substitute with units, then calculate. Before you substitute, ask: is this the RESULTANT force, and is the mass in kg?

Try it: in PhET: Forces and Motion: Basics (version 2.6.0), open the Acceleration screen, apply the same force to a crate and then to a crate plus a person, and compare the accelerations.

Newton's second law of motion (Khan Academy). Watch how the net (resultant) force, not any single force, is used in F = ma, and how the units give the newton.

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Practice

State Newton's second law of motion.
Define the newton.
Find the resultant force needed to give a 5.0 kg box an acceleration of $3.0\,\mathrm{m\,s^{-2}}$ .
A resultant force of 12 N acts on a 400 g ball. Find its acceleration.
A car of mass 900 kg has a driving force of 2500 N and a resistive force of 700 N. Find its acceleration.
A resultant force of 600 N gives a motorcycle and rider an acceleration of $4.0\,\mathrm{m\,s^{-2}}$ . What is their total mass?
One elastic band gives a trolley an acceleration of $0.80\,\mathrm{m\,s^{-2}}$ . Predict the acceleration (a) with three identical bands, (b) with one band and twice the mass.
Use the second law to explain why an overloaded lorry is dangerous on a long downhill road.

Show answersHide answers

The acceleration of a body is directly proportional to the resultant force on it and inversely proportional to its mass, in the direction of the resultant force: $F = m a$ .
One newton is the resultant force that gives a 1 kg mass an acceleration of $1\,\mathrm{m\,s^{-2}}$ .
$F = 5.0 \times 3.0 = 15\,\mathrm{N}$
$m = 0.40\,\mathrm{kg}$ ; $a = \frac{12}{0.40} = 30\,\mathrm{m\,s^{-2}}$
Resultant $= 2500 - 700 = 1800\,\mathrm{N}$ ; $a = \frac{1800}{900} = 2.0\,\mathrm{m\,s^{-2}}$
$m = \frac{F}{a} = \frac{600}{4.0} = 150\,\mathrm{kg}$
(a) three times the force: $2.4\,\mathrm{m\,s^{-2}}$ . (b) twice the mass: $0.40\,\mathrm{m\,s^{-2}}$ .
The deceleration is $a = \frac{F}{m}$ . The brakes can give only a limited force, so a larger mass gives a smaller deceleration: the lorry takes longer and further to stop, and may not be able to stop at all.

Key points

Second law: $F = m a$ , where $F$ is the RESULTANT force in N, $m$ the mass in kg and $a$ the acceleration in $\mathrm{m\,s^{-2}}$ .
$a$ is proportional to $F$ (fixed mass) and inversely proportional to $m$ (fixed force).
$1\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}$ : the force that gives 1 kg an acceleration of $1\,\mathrm{m\,s^{-2}}$ .
A larger mass needs a larger force for the same acceleration or deceleration.

Sources and credits

4.3: Newton's Second Law of Motion: Concept of a System, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Photo: Tours du Rwanda in Musanze.jpg by Muhirehussein12, Wikimedia Commons, licensed under CC BY-SA 4.0.
Forces and Motion: Basics (version 2.6.0). Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY 4.0 (https://phet.colorado.edu)
Video: Newton's second law of motion | Forces and Newton's laws of motion | Physics | Khan Academy, Khan Academy, embedded from YouTube.

Written by Ganzaa: the lesson text, the experiment, results tables, graph, diagrams, worked examples, road-safety application and practice exercises.

1Same engine, different load

2Experiment: how acceleration depends on force and on mass

3Sample results

4From the results to the law

5The second law on the road

Same engine, different load

Experiment: how acceleration depends on force and on mass

Sample results

From the results to the law

The second law on the road