Physics · 1 Motion, forces and energy

1.5.1 Effects of forces

Before you start

A force is a push or a pull, measured in newtons (N), and it has a direction.
Weight is the force of gravity on a mass: in the exam, a mass of $1.0$ kg weighs $9.8$ N.
Velocity is speed in a stated direction, and acceleration is the change in velocity per unit time, in m/s².
Converting lengths: $100$ cm $= 1$ m, so $5.0$ cm $= 0.050$ m.
Rearranging a three-letter equation such as $F = m a$ to make any one letter the subject.

Syllabus objectives covered (12)

Know that forces may produce changes in the size and shape of an object
Sketch, plot and interpret load–extension graphs for an elastic solid and describe the associated experimental procedures
Determine the resultant of two or more forces acting along the same straight line
Know that an object either remains at rest or continues in a straight line at constant speed unless acted on by a resultant force
State that a resultant force may change the velocity of an object by changing its direction of motion or its speed
Describe solid friction as the force between two surfaces that may impede motion and produce heating
Know that friction (drag) acts on an object moving through a liquid
Know that friction (drag) acts on an object moving through a gas (e.g. air resistance)
Define the spring constant as force per unit extension; recall and use the equation k = F / x
Define and use the term ‘limit of proportionality’ for a load–extension graph and identify this point on the graph (an understanding of the elastic limit is not required)
Recall and use the equation F = ma and know that the force and the acceleration are in the same direction
Describe, qualitatively, motion in a circular path due to a force perpendicular to the motion as: (a) speed increases if force increases, with mass and radius constant (b) radius decreases if force increases, with mass and speed constant (c) an increased mass requires an increased force to keep speed and radius constant (F = mv2 / r is not required)

Part 1 of 7

Two things a force can do

When a force acts on an object, it can do one of two things, or both at once. It can change how the object moves. Or it can change the object itself.
Changing the object itself means changing its size or its shape. Hang a load on a spring and the spring gets longer: that is a change in size. Bend a plastic ruler and its shape changes. Squash a sponge and both its size and its shape change.
Often both effects happen together. A tennis ball squashes against the racket for a moment, and in that same moment its direction of motion reverses.
The syllabus wording is 'changes in the size and shape'. If a question asks what a force can do to an object, use those two words, and add 'change its speed or direction' for the motion effect.
Some materials go back to their original size and shape when the force is removed, like a spring stretched gently. These are called elastic. The rest of this lesson uses a spring as the example of an elastic solid, because its size change is easy to measure.

Part 2 of 7

The spring experiment: measuring extension

The quantity you measure is extension, not length. Extension is how much longer the spring is than it was with nothing hanging on it.
$extension = stretched length - original length$
Set up the apparatus. Hang the spring from a clamp on a stand. Clamp a metre rule vertically beside it. Fix a small pointer to the bottom of the spring so that it lies across the ruler's scale.
The pointer turns the bottom of the spring into a reading on the ruler. Your eye must be level with it.
With no load on the spring, read the position of the pointer on the ruler. This is your original reading. Every extension is measured from it.
Add a known load. The load is the weight of the masses you hang on the spring. With $1.0$ kg weighing $9.8$ N, each $100$ g mass adds $0.98$ N of load.
Wait until the spring stops bouncing. Then read the pointer with your eye level with it. If you look from above or below, the pointer seems to sit against a different mark on the scale. That is a parallax error. Reading at eye level avoids it, and it is a point worth stating whenever you describe this experiment.
Calculate the extension for this load by subtracting the original reading. Then add the next mass and repeat, until you have at least five or six pairs of load and extension readings.
Repeat the whole set of readings and take a mean for each load. This makes the results more reliable. For safety, put a soft pad on the bench under the masses and keep your feet clear, in case the hanger falls.
Plot the results with a line of best fit. The question tells you which quantity goes on which axis. Read the axis labels before you plot or sketch, because the two choices give graphs of different steepness.

Part 3 of 7

Reading the graph: spring constant and limit of proportionality

For a spring, the first part of the load–extension graph is a straight line through the origin. That tells you that extension is directly proportional to load: double the load and the extension doubles.
In that straight part, the force needed for each metre of stretch is the same at every point. That fixed value is the spring constant, $k$ . Its definition is force per unit extension. Use those words when you are asked to define it.
$k = \frac{F}{x}$
$F$ is the force (the load) in N and $x$ is the extension in m, so $k$ is in N/m. If you leave $x$ in cm, $k$ comes out in N/cm, which is $100$ times smaller than the value in N/m. When a question asks for N/m, convert the extension to metres first.
A large $k$ means a stiff spring: it needs a large force for each metre of stretch. A small $k$ means a spring that stretches easily.
There is no formula sheet in Paper 4, so you must recall $k = F / x$ and rearrange it yourself. Multiply both sides by $x$ to get $F = k x$ . Then divide both sides by $k$ to get $x = F / k$ .
The gradient of the straight part depends on the axes. With load on the y-axis and extension on the x-axis, the gradient is $F / x$ , which is $k$ . With extension on the y-axis and load on the x-axis, the gradient is $x / F$ , which is $1 / k$ . Check the axes before you use a gradient.
At a certain load, the graph stops being a straight line. That point is the limit of proportionality. Beyond it, extension is no longer directly proportional to load.
Straight through the origin up to the limit of proportionality, then curving. Only the straight part gives a value of k.
Beyond the limit of proportionality, a spring stretches more for each extra newton than it did before. So on a graph with extension on the y-axis, the line curves upwards and gets steeper. With load on the y-axis, it bends over towards the extension axis instead.
In a results table, you find the limit of proportionality without a graph. Look at how much the extension increases for each equal step of load. Where that increase stops being constant, you have passed the limit of proportionality.
Use $k = F / x$ only with readings at or below the limit of proportionality. Beyond it, $F / x$ gives a different value at each point, so no single spring constant describes the spring.
You may read about an 'elastic limit' in textbooks. The syllabus does not require it, so do not use it. If a question says 'limit of proportionality', answer about proportionality: where the straight line ends.

Part 4 of 7

Adding forces along one line: the resultant

Usually several forces act on an object at once. The resultant force is the single force that has the same effect as all of them together. It is the resultant, not any one force, that decides what happens to the motion.
When all the forces act along the same straight line, choose one direction as positive. Forces in that direction are positive and forces in the opposite direction are negative. Then add them up, with their signs.
In practice: forces in the same direction add. Forces in opposite directions subtract, and the resultant points the same way as the larger side. For example, $30$ N to the right and $12$ N to the left give a resultant of $30 - 12 = 18$ N to the right.
Two opposite forces replaced by one resultant: 30 N − 12 N = 18 N, towards the larger force.
With three or more forces, total each direction first. Then subtract the smaller total from the larger total. This avoids sign errors.
Always give the direction with the size. ' $18$ N' is half an answer; ' $18$ N to the right' is the whole answer.
If the forces in each direction are equal, the resultant is zero. The forces are then balanced. The object still has forces on it, but their total effect is nothing.

Part 5 of 7

What a resultant force does to motion

Start with the case where there is no resultant force. An object either stays at rest, or continues in a straight line at constant speed. This is Newton's first law.
The second half of that law is the one people get wrong. A moving object does not need a force to keep it moving. It needs a resultant force only to change its motion. In everyday life friction always acts, which is why things seem to need a push to keep going.
Zero resultant force does not mean that no forces act. A car moving at a steady speed on a straight road has a large forward driving force and an equally large backward force from friction and air resistance. They are balanced, so the motion does not change.
When there is a resultant force, the velocity changes. It can change in two ways: the speed changes (the object speeds up or slows down), or the direction of motion changes. Sometimes both change together.
How much the velocity changes per second, the acceleration, depends on the resultant force and the mass. Resultant force equals mass times acceleration.
$F = m a$
$F$ is in N, $m$ in kg and $a$ in m/s². One newton is the force that gives a mass of $1$ kg an acceleration of $1$ m/s². Rearranged: $a = F / m$ and $m = F / a$ .
The $F$ in $F = m a$ is the resultant force. If a question gives you several forces, find the resultant first, then put it into $F = m a$ . Putting in the engine force alone is a very common lost mark.
The acceleration is always in the same direction as the resultant force. If the resultant force on a moving car points backwards, the acceleration points backwards too, so the car slows down. A negative acceleration in your working means 'opposite to the direction you chose as positive'.

Part 6 of 7

Friction and drag

Solid friction is the force between two surfaces that may impede motion. It acts where the surfaces touch, and it opposes them sliding over each other. A box pushed across a carpet has friction on it, acting backwards.
Friction also produces heating. Rub your hands together and they get warm. A car's brakes get hot when it stops. When asked for the effects of friction, give both: it opposes the motion, and it heats the surfaces.
Friction also acts on an object moving through a liquid. This kind of friction is called drag. A swimmer, a boat and a ball sinking through oil all have drag on them, acting opposite to their motion through the liquid.
Drag also acts on an object moving through a gas. Drag in air is called air resistance. A cyclist, a falling parachutist and a moving car all have air resistance on them.
Drag gets larger as the object moves faster through the liquid or gas. That is why a car needs a bigger driving force to travel steadily at high speed than at low speed.
In a resultant-force question, friction, drag and air resistance are the forces that point backwards. Subtract them from the forward force before you use $F = m a$ .
Friction is not always a nuisance. Friction between tyres and the road is what lets a car grip the road, start, stop and turn a corner.

Part 7 of 7

Moving in a circle

Think about an object moving in a circle at constant speed. Its direction of motion changes all the time. A change of direction is a change of velocity, so the first law tells you that there must be a resultant force on it.
That force acts perpendicular to the motion, towards the centre of the circle. At every instant the object moves along the edge of the circle, and the force pulls in at right angles to that.
The force is at right angles to the velocity and points to the centre, so it turns the ball without speeding it up.
Because the force is at right angles to the motion, it does not speed the object up or slow it down. It only changes the direction. This is how speed stays constant while velocity changes.
The force can come from different places. For a ball whirled on a string, it is the pull (tension) of the string. For a car going round a roundabout, it is friction between the tyres and the road. For the Moon going round the Earth, it is gravity.
If the force disappears, for example if the string breaks, there is no resultant force. The ball then moves off in a straight line at constant speed, in the direction it was travelling at that instant. It does not fly outwards from the centre.
The syllabus asks for three comparisons. In each one, two things stay the same and you say what happens to the third. First: with the same mass and the same radius, a larger force goes with a greater speed. Whirl a ball faster on the same length of string and you have to pull harder.
Second: with the same mass and the same speed, a larger force makes the radius smaller. A bigger inward pull turns the object more sharply, into a tighter circle. Less force means a wider circle.
Third: with the same speed and the same radius, a larger mass needs a larger force. A loaded lorry needs more friction from its tyres than a small car to go round the same roundabout at the same speed.
You need no equation for this. The syllabus says these descriptions are qualitative, so answer in words: 'the force must increase', 'the radius decreases'. Always say which quantities you are keeping the same.

Worked examples

Example 1. Two people push a crate across a floor. One pushes to the right with $120$ N and the other pushes to the right with $90$ N. Friction on the crate is $150$ N to the left. (a) Determine the resultant force on the crate. (b) The moving crate then reaches a rougher part of the floor, where friction is $210$ N. The pushes stay the same. Describe the motion of the crate on this part of the floor.

Show the working

Take right as positive. Both pushes are positive and friction is negative, because it acts to the left.
Add the forces with their signs.
$F = 120 + 90 - 150 = + 60 N$
The answer is positive, so the resultant is $60$ N to the right. While this resultant acts, the crate speeds up to the right.
On the rough floor, do the same sum with the new friction force.
$F = 120 + 90 - 210 = 0 N$
The resultant force is zero. The first law says the crate's motion does not change. It keeps moving in a straight line at the speed it had when it reached the rough floor.
The trap here is to say the crate stops. Slowing down needs a resultant force acting backwards, and there is none.

Answer: (a) $60$ N to the right. (b) The resultant force is zero, so the crate continues in a straight line at constant speed.

Example 2. A car of mass $1400$ kg moves along a straight road. The driver brakes, and the resultant force on the car is $5600$ N acting backwards. (a) Calculate the acceleration of the car and state its direction. (b) Later the car speeds up with an acceleration of $2.5$ m/s². Calculate the resultant force on the car.

Show the working

Recall $F = m a$ . You want $a$ , and $m$ is multiplying it, so divide both sides by $m$ .
$a = \frac{F}{m} = \frac{5600}{1400} = 4.0 {m/s}^{2}$
Now the direction. The acceleration is in the same direction as the resultant force, so it is backwards. A backwards acceleration on a car that moves forwards means that the car slows down.
For (b), use $F = m a$ as it stands, because you have both $m$ and $a$ .
$F = 1400 \times 2.5 = 3500 N$
This is the resultant force, forwards. The engine must give more than $3500$ N, because friction and air resistance act backwards as well. The question asks only for the resultant.

Answer: (a) $4.0$ m/s² backwards: the car decelerates. (b) $3500$ N forwards.

Example 3. A spring has an unstretched length of $15.0$ cm. When a load of $4.0$ N hangs from it, its length is $20.0$ cm. (a) Calculate the spring constant in N/m. (b) The limit of proportionality is not passed. Calculate the length of the spring when the load is $6.0$ N.

Show the working

Find the extension first, not the length. Then convert it to metres, because the answer must be in N/m.
$x = 20.0 - 15.0 = 5.0 cm = 0.050 m$
Use the definition: force per unit extension.
$k = \frac{F}{x} = \frac{4.0}{0.050} = 80 N/m$
For (b), you want $x$ . Multiply both sides of $k = F / x$ by $x$ , then divide both sides by $k$ .
$x = \frac{F}{k} = \frac{6.0}{80} = 0.075 m = 7.5 cm$
The question asks for the length, not the extension. Add the original length back on.
$15.0 + 7.5 = 22.5 cm$
Check by proportion: $6.0$ N is $1.5$ times $4.0$ N, so the extension must be $1.5 \times 5.0 = 7.5$ cm. It is.

Answer: (a) $80$ N/m. (b) $22.5$ cm.

Example 4. A student loads a spring and records these results. Load / N: $0$ , $1.0$ , $2.0$ , $3.0$ , $4.0$ , $5.0$ , $6.0$ . Extension / cm: $0$ , $2.0$ , $4.0$ , $6.0$ , $8.0$ , $10.6$ , $14.0$ . (a) Identify the load at the limit of proportionality. (b) Calculate the spring constant in N/m. (c) Describe the shape of a graph of extension (y-axis) against load (x-axis) for these results.

Show the working

Work out how much the extension increases for each $1.0$ N step of load: $2.0$ , $2.0$ , $2.0$ , $2.0$ , then $2.6$ , then $3.4$ cm.
Up to $4.0$ N the increase is the same every time, so extension is proportional to load. After $4.0$ N the increase is larger each time, so it is not. The limit of proportionality is at about $4.0$ N. The data show only that it is between $4.0$ N and $5.0$ N.
For (b), use a reading in the proportional part. At $4.0$ N the extension is $8.0$ cm, which is $0.080$ m.
$k = \frac{F}{x} = \frac{4.0}{0.080} = 50 N/m$
Any reading up to $4.0$ N gives the same value, for example $1.0 / 0.020 = 50$ N/m. A reading beyond the limit gives a different value, for example $6.0 / 0.140 = 43$ N/m. That is why you must not use it.
For (c): the points up to $4.0$ N lie on a straight line through the origin. Beyond $4.0$ N each newton adds more extension than before, so the line curves upwards and gets steeper.

Answer: (a) About $4.0$ N. (b) $50$ N/m. (c) A straight line through the origin up to $4.0$ N, then a curve that gets steeper.

Where people go wrong

Mistake: Putting the stretched length into $k = F / x$ instead of the extension, or leaving the extension in cm when the answer is wanted in N/m.

Instead: Subtract the original length first, then convert cm to m by dividing by $100$ .

$x$ is the extension, the increase in length. Using the length gives a value that is too small, and using cm for N/m gives a value $100$ times too small.

Mistake: Saying that an object moving at constant speed must have a resultant force forwards to keep it going.

Instead: Constant speed in a straight line means the resultant force is zero. The forward and backward forces are balanced.

A resultant force changes velocity. If the velocity is not changing, there is no resultant force. Everyday experience misleads you because friction is always there.

Mistake: Writing 'no forces act' when the resultant force is zero.

Instead: Write 'the forces are balanced' or 'the resultant force is zero'.

A car at steady speed has large forces on it. They cancel. 'No forces' is wrong science and an examiner will not credit it.

Mistake: Thinking that an object moving in a circle at constant speed is not accelerating, or that the force on it acts outwards.

Instead: Its velocity changes because its direction changes, so it accelerates. The resultant force acts towards the centre, perpendicular to the motion.

Velocity has a direction. If the force were removed the object would move off in a straight line along its direction of travel, not outwards.

Mistake: Using one of the forces, such as the engine force, in $F = m a$ .

Instead: Find the resultant force first, then use $F = m a$ .

Acceleration depends on the resultant force. Friction and drag act backwards and reduce it.

Check yourself

Apart from changing its motion, what two changes can a force produce in an object? Give an example of each.
Answer
A change in size, for example a spring getting longer when it is stretched. A change in shape, for example a ruler bending.
Three forces act on an object along one line: $8.0$ N to the left, $5.0$ N to the right and $2.0$ N to the left. Determine the resultant force.
Answer
Total to the left $= 8.0 + 2.0 = 10.0$ N. Resultant $= 10.0 - 5.0 = 5.0$ N to the left.
A puck slides across smooth ice at constant speed in a straight line. A friend says there must be a forward force to keep it moving. Is your friend right? Then state the two ways in which a resultant force could change the puck's velocity.
Answer
No. Constant speed in a straight line means the resultant force is zero, and no force is needed to keep it moving. A resultant force could change its speed or change its direction of motion.
A swimmer, a falling parachutist and a box sliding across a carpet are each slowed by a type of friction. Name the type for each. Then give one effect of the friction on the box apart from slowing it.
Answer
Swimmer: drag (from the water, a liquid). Parachutist: air resistance (drag from a gas). Box: solid friction. The friction also heats the box and the carpet.
A spring has a spring constant of $40$ N/m. Calculate its extension when a load of $2.0$ N hangs from it, assuming the limit of proportionality is not passed.
Answer
$x = F / k = 2.0 / 40 = 0.050$ m, which is $5.0$ cm.
What is the limit of proportionality, and how do you find it on a load–extension graph?
Answer
It is the point beyond which extension is no longer directly proportional to load. On the graph it is the point where the straight line through the origin ends and the line starts to curve.
A ball of mass $0.25$ kg has an acceleration of $20$ m/s² to the left. Calculate the resultant force on it and state its direction.
Answer
$F = m a = 0.25 \times 20 = 5.0$ N, to the left, in the same direction as the acceleration.
A ball on a string moves in a horizontal circle at constant speed. It is replaced by a ball of greater mass, moving at the same speed in a circle of the same radius. What must happen to the pull of the string? And what happens to the radius if the pull increases while the mass and speed stay the same?
Answer
The pull must increase, because a larger mass needs a larger force for the same speed and radius. If the pull increases with mass and speed constant, the radius decreases.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student investigates how a spring stretches. The unstretched length of the spring is $20.0$ cm. She hangs loads on the spring and records its length each time. Load / N: $0$ , $0.50$ , $1.00$ , $1.50$ , $2.00$ , $2.50$ , $3.00$ . Length / cm: $20.0$ , $22.5$ , $25.0$ , $27.5$ , $30.0$ , $33.0$ , $37.0$ .

(a) Describe how the student obtains the extension of the spring for each load. 3 marks No calculator

Mark scheme

B1 measures the length with no load (original length) using a metre rule held vertically beside the spring
B1 adds the loads one at a time and measures the new length each time; also credit reading the scale at eye level with the pointer to avoid parallax, or waiting until the spring stops moving
B1 extension = new length − original length

(b) Calculate the spring constant of the spring, in N/m. 3 marks

Mark scheme

C1 extension for a load of 2.00 N or less, e.g. 30.0 − 20.0 = 10.0 cm for 2.00 N
C1 k = F / x with the extension in m, e.g. 2.00 / 0.100
A1 20 N/m; unit required

(c) Identify the load at which the limit of proportionality is reached. Use the results to justify your answer. 2 marks No calculator

Mark scheme

B1 2.00 N (accept any value between 2.00 N and 2.50 N)
B1 up to 2.00 N the extension increases by 2.5 cm for each 0.50 N, but from 2.00 N to 2.50 N it increases by 3.0 cm / extension is no longer proportional to load beyond this point, owtte

Question 2. A motor boat of mass $800$ kg moves in a straight line across a lake. The engine gives a forward force of $2400$ N. The water exerts a backward force of $1000$ N on the boat and the air exerts a backward force of $200$ N.

(a) State the name given to the backward forces that the water and the air exert on the moving boat. 1 mark No calculator

Mark scheme

B1 drag / friction (accept air resistance for the force from the air)

(b) Determine the resultant force on the boat. 2 marks

Mark scheme

C1 2400 − 1000 − 200 or 2400 − 1200 seen
A1 1200 N forwards; unit required

(c) Calculate the acceleration of the boat. 2 marks

Mark scheme

C1 a = F / m or 1200 / 800, ecf from (b)
A1 1.5 m/s²; unit required

(d) Later, the boat moves in a straight line at a constant speed. The engine force is still $2400$ N. Deduce the total backward force on the boat now, and give a reason for your answer. 2 marks No calculator

Mark scheme

B1 2400 N
B1 the speed is constant / there is no acceleration, so the resultant force is zero / the forces are balanced

Question 3. A car drives round a circular roundabout at a constant speed. The force that keeps the car on its circular path is friction between the tyres and the road.

(a) State the direction of this friction force. 1 mark No calculator

Mark scheme

B1 towards the centre of the circle (perpendicular to the direction of motion)

(b) Explain why the car is accelerating, even though its speed is constant. 2 marks No calculator

Mark scheme

B1 the direction of motion of the car is changing
B1 so its velocity is changing / there is a resultant force on it

(c) A loaded lorry with a much greater mass drives round the same roundabout at the same speed. Compare the force needed to keep the lorry on the circular path with the force needed for the car. 1 mark No calculator

Mark scheme

B1 the lorry needs a larger force, ORA

(d) On an icy day the friction between the tyres and the road is much smaller. The car enters the roundabout at the same speed. Predict what happens to the radius of the car's path, and give a reason. 2 marks No calculator

Mark scheme

B1 the radius increases / the car moves in a wider curve / the car skids towards the outside of the roundabout
B1 smaller force with the same mass and speed gives a larger radius, ORA

How this is examined

How you measure the extension of a spring for each load: a three- or four-mark Describe, often before a results table that the experiment produced Measure the original length with no load; add known loads one at a time; measure the new length with a ruler beside the spring, eye level with the pointer to avoid parallax; extension = new length − original length. Each of these is a separate mark.
Getting the spring constant from a table or a graph: a two- or three-mark Calculate after a results table, sometimes asking for the answer in N/m when the data are in cm Extension (not length) taken from a point in the straight part; extension converted to m; $k = F / x$ written; the answer with the unit N/m.
Resultant force first, then $F = m a$ : a two-part calculation: Determine the resultant force from a forward force and some backward forces, then Calculate the acceleration The backward forces subtracted from the forward force, with the direction stated; then $a = F / m$ using that resultant, with the unit m/s². An error in the resultant still earns the second mark through ecf.
Why something in a circle at constant speed is accelerating, and what changes the force: a two-mark Explain followed by one-mark Predict or Compare parts that change the mass, the speed or the radius 'The direction of motion changes, so the velocity changes'; 'the force is towards the centre'; then 'a larger force is needed' for more mass or more speed, or 'the radius decreases' for more force, with the other two quantities stated as unchanged.
Balanced forces mean constant velocity, not 'stopped': a one- or two-mark State or Deduce about an object moving at constant speed, often asking you to find the size of an unknown backward force 'The resultant force is zero' (not 'there are no forces'), and so the unknown backward force equals the forward force.
Finding the limit of proportionality: a one- or two-mark Identify on a load–extension graph or table, or a one-mark Define The load where the straight line ends (or where the extension per step stops being constant), with the reason 'extension is no longer proportional to load'.

Words to know

Load: The force that stretches a spring, usually the weight of the masses hung from it. Measured in N.
Extension: The increase in length of a spring: stretched length − original length.
Spring constant: Force per unit extension, $k = F / x$ . Measured in N/m.
Limit of proportionality: The point on a load–extension graph beyond which extension is no longer directly proportional to load; where the straight line through the origin ends.
Resultant force: The single force that has the same effect as all the forces acting on an object together.
Solid friction: The force between two surfaces that may impede motion and produce heating.
Drag: Friction on an object moving through a liquid or a gas. Drag in air is called air resistance.
Newton (N): The unit of force: the force that gives a mass of $1$ kg an acceleration of $1$ m/s².

All Physics lessons Next: 1.5.2 Turning effect of forces