Physics · 1 Motion, forces and energy

1.7.2 Work

Before you start

Forces are measured in newtons (N), and weight is a force: on Paper 4 you take the weight of 1.0 kg to be 9.8 N.
Energy stores from 1.7.1: kinetic, gravitational potential, chemical, elastic, internal (thermal), all measured in joules (J).
The principle of conservation of energy: energy is transferred between stores, never created or destroyed.
With balanced forces (zero resultant force) an object moves at constant speed in a straight line.
Extended, from 1.7.1: $E_{k} = \frac{1}{2} m v^{2}$ and $Δ E_{p} = m g Δ h$ .
Converting cm and km to m, and rearranging an equation of the form $a = b c$ .

Syllabus objectives covered (2)

Understand that mechanical or electrical work done is equal to the energy transferred
Recall and use the equation for mechanical working W = Fd = ΔE

Part 1 of 5

Work is energy transferred by a force

In everyday speech, 'work' means effort. In physics it means something narrower that you can measure: work is done when a force moves an object in the direction of the force.
Doing work transfers energy. Push a trolley so that it speeds up, and energy leaves the chemical store in your muscles and arrives in the kinetic store of the trolley. The push is the route the energy took.
That gives the central idea of this subtopic: the work done is equal to the energy transferred. If a force does $200$ J of work, then $200$ J of energy has been transferred. Not roughly $200$ J. Exactly $200$ J.
This is why work and energy have the same unit, the joule (J). Work done is not a separate energy store. It is the amount of energy that a force moved from one store to another, or from one object to another.
A transfer done by a force is called mechanical work. Lifting, pushing, pulling, stretching a spring and braking are all mechanical work.
The energy does not always go where you want it to go. Push a heavy box across a carpet at a steady speed: its kinetic store does not increase at all. The work you do against friction goes to the internal (thermal) store of the box, the carpet and the air, which all warm up a little. The work done still equals the energy transferred. The energy just went somewhere less useful.

Part 2 of 5

Electrical work: the same rule, done by a current

A force is not the only way to do work. Energy is also transferred by an electric current. When a current passes through a component, the energy it transfers to that component is called electrical work done.
The rule does not change: electrical work done is equal to the energy transferred. If the current in a heater does $3000$ J of electrical work, $3000$ J of energy has been transferred to the heater, which transfers it to the room by heating.
Follow a cell and a lamp. The energy starts in the chemical store of the cell. The current does electrical work on the lamp. The lamp then transfers that energy to the surroundings by light and by heating.
Chains of transfers are where the two kinds of work meet. In an electric motor that lifts a load, the current does electrical work on the motor, and the motor does mechanical work on the load, which gains gravitational potential energy. At each link in the chain, work done equals energy transferred.
The mechanical work the motor does is less than the electrical work done on it. Some of the energy goes to the internal (thermal) store of the motor and its surroundings, because the motor warms up, and some is transferred away as sound. No energy disappears. It is just not all useful.
For this subtopic you need the idea of electrical work, not an equation for it. You calculate electrical energy later, in 4.2.5, with $E = I V t$ .

Part 3 of 5

Calculating mechanical work: W = Fd

The work done by a force is the force multiplied by the distance moved in the direction of the force.
$W = F d$
$W$ is the work done in joules (J). $F$ is the force in newtons (N). $d$ is the distance moved in the direction of the force, in metres (m). Every word of that last phrase matters, and the next section shows why.
Paper 4 gives you no formula sheet, so you must recall $W = F d$ and know what each symbol means. On Papers 5 and 6, a formula you need is given in the question.
The unit comes from the equation: newtons multiplied by metres. So $1$ J $= 1$ N m. One joule is the work done when a force of $1$ N moves an object $1$ m in the direction of the force.
A $40$ N push that moves a box $3.0$ m along the floor does $40 \times 3.0 = 120$ J of work, and so transfers $120$ J of energy.
Change the distance to metres before you multiply. The same $40$ N push over $60$ cm does $40 \times 0.60 = 24$ J, not $40 \times 60 = 2400$ J. A length in centimetres gives an answer $100$ times too big.
When the work is given and you need the force or the distance, rearrange. To find $d$ , divide both sides by $F$ , because $F$ is what multiplies $d$ . To find $F$ , divide both sides by $d$ .
$d = \frac{W}{F} F = \frac{W}{d}$
Large amounts of work are given in kilojoules: $1$ kJ $= 1000$ J. Change kJ to J before you use the value in $W = F d$ , so that the force comes out in N and the distance in m.

Part 4 of 5

The distance must be in the direction of the force

A force does work only when the object moves along the line of that force. Movement in some other direction does not count.
No movement means no work. Push hard against a wall that does not move: $d = 0$ , so $W = 0$ . Hold a heavy bag still: the bag does not move, so you do no work on it. You still get tired, because your muscles transfer energy inside your body, but none of that energy goes to the bag.
Movement at right angles to a force means that force does no work. Carry a bag across a level room at a constant height. Your hand pulls the bag upwards, but the bag moves sideways. None of its movement is in the direction of your upward force, so that force does no work on the bag.
forces diagram — not drawn yet
A shopping bag carried horizontally across a level floor: a vertical upward arrow on the bag labelled 'force from hand', a vertical downward arrow labelled 'weight', and a separate horizontal arrow underneath labelled 'direction of motion'.
The force from the hand is at right angles to the motion, so it does no work on the bag.
Gravity follows the same rule. A ball rolls along a level floor. Its weight acts downwards, the ball moves sideways, so gravity does no work on it, and the ball's gravitational potential energy does not change.
Lifting is the opposite case. Lift a bag straight up and the bag moves in the direction of your upward force, so the work you do is the force multiplied by the vertical height.
For work done against gravity, the distance is the vertical height gained, not the length of the path. Carry a box up a staircase and the work done against gravity is its weight multiplied by the height of the stairs, not the length of the staircase.

Part 5 of 5

W = Fd = ΔE: from force to energy and back

The full equation in the syllabus links the two ideas of this subtopic.
$W = F d = Δ E$
$Δ E$ is the energy transferred. Read the equation in both directions. From force and distance you can find the energy transferred. From the energy transferred you can find a force or a distance.
Lifting at a steady speed is the simplest case. At a steady speed the resultant force is zero, so the upward force equals the weight. The work done is weight × height, and all of it goes to the gravitational potential store. Because weight $= m g$ , this is the same as $Δ E_{p} = m g Δ h$ from 1.7.1.
$F$ must be a force in newtons. If a question gives a mass in kg, change it to a weight first: on Paper 4, every $1.0$ kg has a weight of $9.8$ N. A mass in kg is never $F$ .
In this topic $W$ means work done. Weight can also be given the symbol $W$ . When a calculation uses both, write them in words — 'work done = weight × height' — so that you do not mix them up.
Braking uses the equation the other way round. The braking force does work on a moving car, and the car's kinetic energy is transferred to the internal (thermal) store of the brakes and the surroundings. So the work done by the brakes equals the kinetic energy the car loses.
$F_{brake} \times d = E_{k}$
For a constant braking force, the braking distance is proportional to the kinetic energy. Double the speed and the kinetic energy becomes four times as large, because $E_{k} = \frac{1}{2} m v^{2}$ . So the braking distance becomes four times as long.
When friction acts, the work done splits into two parts. Part goes to the useful store, and the rest goes to the internal (thermal) store by heating. Work done against friction = total work done − useful energy gained. Worked example 3 does this for a ramp.
forces diagram — not drawn yet
A box being pushed up a straight ramp 5.0 m long whose top is 1.5 m above the floor: an arrow along the ramp surface labelled 'push 80 N', a vertical downward arrow labelled 'weight 200 N', the sloping length labelled 5.0 m and the vertical height labelled 1.5 m.
The push acts over 5.0 m along the ramp, but the box gains only 1.5 m of height. Each distance answers a different question.

Worked examples

Example 1. A shop worker pushes a loaded trolley along an aisle with a horizontal force of $45$ N. The trolley moves $12$ m in the direction of the push. Calculate the work done by the push and state the energy transferred. How far must she push the trolley with the same force to do $900$ J of work?

Show the working

Write the equation first. It earns credit even if the arithmetic goes wrong later.
$W = F d$
The force is already in N and the distance is already in m, and the distance is in the direction of the force. So substitute directly.
$W = 45 \times 12 = 540 J$
Work done equals energy transferred, so $540$ J of energy is transferred by the push.
For the second part, the unknown is $d$ . Divide both sides of $W = F d$ by $F$ , because $F$ is what multiplies $d$ .
$d = \frac{W}{F} = \frac{900}{45} = 20 m$
Check the size: $900$ J is a little less than twice $540$ J, and $20$ m is a little less than twice $12$ m. The answer is reasonable.

Answer: Work done $= 540$ J, so $540$ J of energy is transferred. To do $900$ J of work she must push it $20$ m.

Example 2. A crane lifts a crate of mass $350$ kg vertically at a steady speed through a height of $18$ m. Calculate the work done by the crane on the crate, and deduce the gain in gravitational potential energy of the crate. (Take the weight of $1.0$ kg to be $9.8$ N.)

Show the working

The equation needs a force, not a mass. Find the weight of the crate first.
$weight = 350 \times 9.8 = 3430 N$
The crate moves at a steady speed, so the resultant force on it is zero. The upward force from the crane is therefore equal to the weight, $3430$ N.
The crate moves vertically upwards, in the direction of that force, so $d$ is the height.
$W = F d = 3430 \times 18 = 61 740 J$
Give the answer to two significant figures to match the data: $62 000$ J, which is $62$ kJ.
The speed is steady, so the kinetic energy does not change. All the work done goes to the gravitational potential store. Work done equals energy transferred, so the gain in gravitational potential energy is also $62 000$ J.

Answer: Work done $= 61 740$ J $\approx 62$ kJ; gain in gravitational potential energy $\approx 62$ kJ.

Example 3. A box of weight $200$ N is pushed at a steady speed up a ramp $5.0$ m long. The top of the ramp is $1.5$ m above the floor. The push is $80$ N, along the ramp. Calculate the work done by the push and the gain in gravitational potential energy of the box. Determine the energy transferred by friction, and the friction force.

Show the working

The push acts along the ramp, and the box moves $5.0$ m along the ramp. So for the push, $d = 5.0$ m.
$W_{push} = 80 \times 5.0 = 400 J$
For the gain in gravitational potential energy, the force is the weight, which acts vertically. The distance in that direction is the vertical height, $1.5$ m, not the length of the ramp.
$Δ E_{p} = 200 \times 1.5 = 300 J$
The speed is steady, so the kinetic energy does not change. Work done equals energy transferred, so the $400$ J must all go somewhere. $300$ J goes to the gravitational potential store, so the rest goes to the internal (thermal) store of the box, the ramp and the surroundings, through friction.
$400 - 300 = 100 J$
Friction acts along the ramp over the same $5.0$ m. Use $W = F d$ again, rearranged for $F$ .
$F = \frac{W}{d} = \frac{100}{5.0} = 20 N$
Notice what the ramp does. If there were no friction, the push would need to do only $300$ J. The ramp lets you use a smaller force than the $200$ N weight, but over a longer distance.

Answer: Work done by the push $= 400$ J; gain in gravitational potential energy $= 300$ J; energy transferred by friction $= 100$ J; friction force $= 20$ N.

Where people go wrong

Mistake: Using the mass in kg as the force, e.g. writing $350 \times 18 = 6300$ J for a $350$ kg crate lifted $18$ m.

Instead: Change mass to weight first ( $350 \times 9.8 = 3430$ N), then use $W = F d$ .

$F$ in $W = F d$ is a force in newtons. A mass in kg is not a force, and the answer comes out about ten times too small.

Mistake: Using the total distance travelled when only part of the motion is in the direction of the force — for example, weight × length of a ramp for the gain in gravitational potential energy.

Instead: Use the distance moved in the direction of the force. For work against gravity, that is the vertical height.

The definition is force × distance moved in the direction of the force. Motion at right angles to a force transfers no energy through that force.

Mistake: Saying you do work on a bag when you hold it still, because holding it is tiring.

Instead: The bag does not move, so $d = 0$ and no work is done on the bag.

Feeling tired shows that energy is transferred inside your body. Work done on the bag needs the bag to move in the direction of your force.

Mistake: Putting a distance in cm straight into $W = F d$ , e.g. $25$ N over $40$ cm $= 1000$ J.

Instead: Change the distance to metres first: $25 \times 0.40 = 10$ J.

$1$ J is $1$ N m. With centimetres the answer is $100$ times too large.

Mistake: Thinking that the energy 'missing' from a motor or a ramp has been used up.

Instead: Name where it went: to the internal (thermal) store of the machine and the surroundings by heating, and sometimes as sound.

Work done always equals energy transferred, and energy is conserved. If the useful energy is less than the work done, the difference is in another store.

Check yourself

State the link between the work done by a force and the energy transferred.
Answer
They are equal. The work done by a force is the energy that the force transfers.
A force of $30$ N pushes a box $4.0$ m in the direction of the force. Calculate the work done.
Answer
$W = F d = 30 \times 4.0 = 120$ J.
A force of $250$ N does $2000$ J of work on a sledge. How far does the sledge move in the direction of the force?
Answer
$d = W / F = 2000 / 250 = 8.0$ m.
A student lifts a $1.2$ kg book from the floor onto a shelf $1.5$ m higher. Calculate the work done against gravity. (Weight of $1.0$ kg $= 9.8$ N.)
Answer
Weight $= 1.2 \times 9.8 = 11.76$ N. $W = 11.76 \times 1.5 = 17.64$ J $\approx 18$ J. This is also the gain in gravitational potential energy of the book.
A ball rolls along a level floor. How much work does its weight do on it? Explain your answer.
Answer
Zero. The weight acts vertically downwards and the ball moves horizontally, so the ball moves no distance in the direction of its weight.
An electric motor lifts a load at a steady speed. The current does $500$ J of electrical work on the motor, and the motor does $350$ J of work on the load. How much energy does the load gain, and what happened to the rest?
Answer
The load gains $350$ J of gravitational potential energy. The other $500 - 350 = 150$ J goes to the internal (thermal) store of the motor and the surroundings by heating, and some as sound.

Exam questions

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

Question 1. A forklift truck lifts a pallet of mass $480$ kg vertically at a constant speed through a height of $2.5$ m. Take the weight of $1.0$ kg to be $9.8$ N.

(a) State the equation that links work done, force and the distance moved in the direction of the force. 1 mark No calculator

Mark scheme

B1 W = Fd or work done = force × distance (moved in the direction of the force), or any correct rearrangement (e.g. F = W / d)

(b) Calculate the weight of the pallet. 2 marks

Mark scheme

C1 480 × 9.8 seen
A1 4700 N (accept 4704 N or 4.7 × 10³ N); unit required

(c) Calculate the work done by the forklift truck on the pallet. 2 marks

Mark scheme

C1 W = Fd used with F = weight from (b) and d = 2.5 m, ecf from (b)
A1 12 000 J (accept 11 760 J, 11 800 J, 1.2 × 10⁴ J, 12 kJ or 11.8 kJ); unit required

(d) Deduce the gain in gravitational potential energy of the pallet. 1 mark No calculator

Mark scheme

B1 same value as (c), e.g. 12 000 J, with unit, ecf from (c)

(e) The forklift truck then holds the pallet at rest at this height for $30$ s. Explain why it does no work on the pallet during this time. 2 marks No calculator

Mark scheme

B1 the pallet does not move / distance moved (in the direction of the force) is zero
B1 so W = Fd = 0 / no energy is transferred to the pallet

Question 2. A car of mass $1200$ kg travels along a level road at $15$ m/s. Its kinetic energy is $1.35 \times 10^{5}$ J. The driver brakes, and a constant braking force of $5400$ N brings the car to rest.

(a) Determine the work done by the braking force in bringing the car to rest. 1 mark No calculator

Mark scheme

B1 1.35 × 10⁵ J (accept 135 000 J or 135 kJ); unit required

(b) Calculate the distance the car travels while it is braking. 3 marks

Mark scheme

C1 W = Fd or d = W / F
C1 135 000 / 5400 seen, ecf from (a)
A1 25 m; unit required

(c) The car travels at $30$ m/s instead and the same braking force is applied. Predict the braking distance. 1 mark

Mark scheme

B1 100 m (four times the answer to (b), ecf from (b))

(d) Explain your answer to (c). 2 marks No calculator

Mark scheme

B1 kinetic energy is four times as large / Ek is proportional to v² / Ek = 5.4 × 10⁵ J
B1 work done by the brakes = kinetic energy lost and the force is the same, so the distance is four times as large

Question 3. On a building site, an electric motor lifts a bucket of sand at a constant speed. The weight of the bucket and sand is $600$ N. The bucket is lifted vertically through $8.0$ m. While it does this, the electric current does $6000$ J of electrical work on the motor.

(a) Calculate the work done by the motor on the bucket. 2 marks

Mark scheme

C1 W = Fd or 600 × 8.0 seen
A1 4800 J; unit required

(b) Determine the energy transferred by the motor that does not go to the bucket. 1 mark

Mark scheme

B1 1200 J (6000 − answer to (a), ecf from (a)); unit required

(c) Explain why the work done by the motor on the bucket is less than the electrical work done on the motor. 2 marks No calculator

Mark scheme

B1 work done = energy transferred / energy is conserved, so the 6000 J must all go to some store
B1 some energy goes to the internal (thermal) store of the motor or surroundings by heating (accept sound); insufficient: 'energy is lost' with no store or process named

How this is examined

Work done from a force and a distance, and back again: a two- or three-mark Calculate; the distance may be in cm or km, or you may be given the work and asked for the force or the distance $W = F d$ written, or the numbers substituted into it, earns a C mark; the distance changed to m; the final answer with the unit J (or N m, or N and m for a rearranged answer). A correct answer with its unit scores full marks without working (unless the question says 'show your working'), but if the number is wrong, only the working can score.
Lifting: the force is the weight and the distance is the vertical height: a Calculate that gives a mass and a height, so you must find the weight with 9.8 N per kg before you find the work done or the gain in gravitational potential energy Weight $=$ mass $\times 9.8$ in N; work done $=$ weight $\times$ vertical height; answer in J. If your weight is wrong but you use it correctly in the next step, ecf still gives you the later marks.
Braking distance from W = Fd = ΔE: a Calculate of a braking distance from a kinetic energy and a braking force, then a Predict and Explain about a change of speed Work done by the brakes $=$ kinetic energy lost; $d = E_{k} / F$ with the answer in m. For the Predict: kinetic energy is four times as large when the speed doubles, the braking force is unchanged, so the braking distance is four times as large.
Where the rest of the energy went when a machine or friction is involved: a Determine (subtract the useful energy from the total work done) followed by a two-mark Explain Total work done − useful energy gained $=$ energy to the internal (thermal) store of the machine or surroundings by heating; 'work done = energy transferred' or 'energy is conserved'. 'Used up' does not name a store, so it does not describe where the energy went.
The energy gained is the same number as the work done: a one- or two-mark State or Deduce straight after you have calculated a work done The same value in J as the work done, carried forward from your earlier answer, with the store named (gravitational potential, kinetic or internal).
Why a force can act and still do no work: a one- or two-mark Explain about an object held still, or an object moving at right angles to a force 'The object does not move in the direction of the force', so 'distance $= 0$ ', so ' $W = F d = 0$ ' or 'no energy is transferred to the object'.

Words to know

Work done: Force × distance moved in the direction of the force, $W = F d$ . It is equal to the energy transferred. Measured in joules.
Mechanical work: Energy transferred by a force moving an object, for example lifting, pushing or braking.
Electrical work: The work done when an electric current transfers energy to a component. Its value in J is equal to the energy transferred.
Joule (J): The unit of work and of energy. $1$ J is the work done when a force of $1$ N moves an object $1$ m in the direction of the force, so $1$ J $= 1$ N m.
ΔE: The energy transferred, or the change in the energy in a store. In $W = F d = Δ E$ it equals the work done.
Weight: The gravitational force on an object, in newtons. On Paper 4 the weight of $1.0$ kg is $9.8$ N.

All Physics lessons Next: 1.7.3 Energy resources