D. Energy and Society · Work, Energy and PowerWork: When Is Work Done?
D3.5 D2.2 D2.1 D3.2 A2.2About 70 minutes
Learning goals- I can describe the conditions needed for work to be done on an object.
- I can calculate work with W = FΔd when the force is along the line of the displacement.
- I can explain why holding, or carrying something horizontally at constant velocity, does no work on it.
- I can tell when work is positive, negative or zero, and explain that work is a transfer of energy.
Words to know
work (W)
The energy transferred to or from an object when a force acts on it while it moves along the line of the force. W = FΔd, measured in joules.
displacement (Δd)
The change in position of an object, a vector measured in metres. In work problems, use the part of the motion along the line of the force.
force (F)
A push or a pull, a vector measured in newtons (N). 1 N = 1 kg·m/s².
joule (J)
1 J = 1 N·m: the work done by a 1 N force acting through a displacement of 1 m along the line of the force.
positive work
Work done by a force in the same direction as the displacement. It adds energy to the object.
negative work
Work done by a force opposite to the displacement, such as friction. It takes energy away from the object.
Hard work... or no work at all?
You hold a heavy backpack at arm's length for a whole minute. Your arm shakes and burns. Surely that is hard work? A physicist would say you did zero work on the backpack. In everyday speech "work" means effort. In physics, work has a precise meaning: it is the energy transferred to an object when a force moves it. If the backpack does not move, no energy goes into it.
This lesson pins down exactly when work is done, and how to calculate it.
The definition of work
When a constant force acts on an object while the object moves along the line of the force, the work done by that force is
W = FΔd
- W is work in joules (J)
- F is the size of the force in newtons (N)
- Δd is the size of the displacement in metres (m), along the line of the force
Because 1 J = 1 N × 1 m, a joule is also a newton-metre (N·m). Work is a scalar, even though force and displacement are vectors. In Grade 8 you wrote this as W = F × d; in Grade 11 we use Δd for displacement because it is the change in position that matters.
Three conditions for work
A force does work on an object only if all three of these are true:
- A force acts on the object.
- The object has a displacement (it moves).
- The force (or part of it) acts along the line of the displacement.
If any one is missing, the work done by that force is zero:
- No displacement. Pushing hard on a brick wall, or holding a backpack still: Δd = 0, so W = 0.
- No force along the motion. A puck gliding across very smooth ice keeps moving with almost no horizontal force on it. Nothing is pushing it forward, so no work is being done on it in that direction.
- Force perpendicular to the motion. You carry a backpack across the room at constant velocity. Your upward force holds it up, but the motion is sideways. The upward force does no work on the backpack.
Positive and negative work
Work is energy being transferred, so its sign tells you which way the energy goes.
- If the force points the same way as the displacement, the work is positive: the force adds energy to the object (for example, your push speeding up a sled).
- If the force points opposite to the displacement, the work is negative: the force takes energy away from the object (for example, friction slowing a sliding hockey bag, or gravity when you throw a ball upward).
So in one dimension: W = +FΔd when F and Δd are in the same direction, and W = −FΔd when they are opposite.
Worked exampleExample 1: Pulling a toboggan
A parent pulls a toboggan across a flat, snowy park with a horizontal rope. The tension in the rope is 45 N, and the toboggan moves 120 m in the direction of the pull. How much work does the rope do on the toboggan?
- The force is along the line of the displacement, so W = FΔd.
- W = (45 N)(120 m)
- W = 5400 J = 5.4 × 10³ J (or 5.4 kJ)
The rope does 5.4 × 10³ J of work on the toboggan.
Worked exampleExample 2: Lifting, then carrying
A student lifts a 12 kg box of books straight up 1.5 m at constant velocity, then carries it 6.0 m across a hallway at constant velocity. Use g = 9.8 N/kg. (a) How much work does the student do on the box while lifting? (b) While carrying?
- (a) At constant velocity the net force is zero, so the student's upward force equals the box's weight: F = mg = (12 kg)(9.8 N/kg) = 117.6 N.
- The force (up) is along the displacement (up): W = FΔd = (117.6 N)(1.5 m) = 176.4 J ≈ 1.8 × 10² J.
- (b) While carrying, the student's force on the box is still up (to hold it), but the displacement is horizontal. The force is perpendicular to the displacement.
- So W = 0 J. (Walking takes effort from the student's body, but no work is done on the box.)
(a) about 1.8 × 10² J; (b) 0 J.
Worked exampleExample 3: Finding the force from the work
Three friends push a stalled car 15 m along a flat road, doing 2.4 kJ of work on it. What average force did they apply along the road?
- Convert: 2.4 kJ = 2400 J.
- Rearrange W = FΔd to F = W ÷ Δd.
- F = 2400 J ÷ 15 m = 160 N
They pushed with an average force of 160 N (1.6 × 10² N) along the road.
Worked exampleExample 4: Negative work by friction
A hockey bag slides 4.0 m across a garage floor before stopping. The friction force on it is 35 N. How much work does friction do on the bag?
- Friction acts opposite to the displacement, so the work is negative.
- W = −FΔd = −(35 N)(4.0 m) = −140 J
- The negative sign means friction removed 140 J of kinetic energy from the bag. That energy became thermal energy in the bag and the floor.
Friction does −140 J (−1.4 × 10² J) of work.
Going further: a force at an angle
SPH3U problems use forces along the line of motion. If a force acts at an angle θ to the displacement (like a rope pulled upward at an angle), only the part of the force along the displacement does work. Then W = FΔd cos θ. This one formula includes all three cases you met today: cos 0° = 1 (force along the motion, W = FΔd), cos 90° = 0 (perpendicular force, W = 0) and cos 180° = −1 (opposite force, W = −FΔd). You will use it often in SPH4U.
Work is an energy transfer
Why is work measured in joules, the same unit as energy? Because doing work is one way of transferring energy. When you do 176 J of work lifting the box, 176 J of energy leaves your body (as chemical energy is used) and is stored in the box as gravitational potential energy. When friction does −140 J of work on the bag, the bag loses 140 J of kinetic energy.
The English physicist James Prescott Joule showed this link in the 1840s. He let falling weights turn a paddle wheel in water and found that a given amount of mechanical work always warmed the water by the same amount. The SI unit of energy is named after him.
Just for funTeacher: "You did no work on your textbook last night."
Student: "I held it above my desk for two hours!"
Teacher: "Exactly. Zero displacement."
Hands-on activityWork or no work?
You need: a full 1 L water bottle (mass about 1.0 kg), a tape measure, a table, and a laundry basket with a few clothes in it.
- Measure the height of the table. Lift the bottle slowly from the floor to the tabletop. Calculate the work you did on it: W = (1.0 kg)(9.8 N/kg)(height in m).
- Carry the bottle slowly across the room at the same height. How much work did your upward force do on it?
- Hold the bottle out in front of you, without moving, for 30 s. How much work did you do on it? Why do your arm muscles still get tired?
- Push the laundry basket across the floor. Which forces on the basket do positive work, negative work and zero work?
- Write one sentence explaining the difference between physics "work" and everyday "effort".
What to look for: for a 0.75 m table, W ≈ (9.8 N)(0.75 m) ≈ 7.4 J. Carrying horizontally and holding still: 0 J done on the bottle. Your muscles still tire because muscle fibres keep contracting and relaxing, transforming chemical energy into thermal energy inside your body. For the basket: your push does positive work, friction does negative work, and gravity and the normal force do zero work.
Your turn
1A curler sweeps with a 30 N force along the path of the broom for 0.50 m. How much work does the force do on the broom head?
W = FΔd = (30 N)(0.50 m) = 15 J.
2How much work is done on a 2.0 kg bag of flour lifted straight up 1.2 m at constant velocity?
F = mg = (2.0)(9.8) = 19.6 N; W = (19.6 N)(1.2 m) = 23.52 J ≈ 24 J.
3A weightlifter holds a 100 kg barbell motionless above their head for 3.0 s. How much work do they do on the barbell during those 3.0 s?
0 J. The barbell has no displacement.
4A server carries a tray horizontally across a restaurant at constant velocity. Does the server's upward force on the tray do work? Explain.
No. The upward force is perpendicular to the horizontal displacement, so W = 0.
5You do 600 J of work pushing a shopping cart 20 m along an aisle. What average force did you apply along the aisle?
F = W ÷ Δd = 600 J ÷ 20 m = 30 N.
6A sled slides 12 m across snow. Friction on it is 18 N. Find the work done by friction, with its sign.
Friction opposes the motion: W = −(18 N)(12 m) = −216 J ≈ −2.2 × 10² J.
7Show that 1 J = 1 kg·m²/s².
1 J = 1 N·m, and 1 N = 1 kg·m/s². So 1 J = (1 kg·m/s²)(1 m) = 1 kg·m²/s².
More practice: warm-up
1Write the formula for work and name each unit.
W = FΔd: W in joules (J), F in newtons (N), Δd in metres (m).
2A force of 10 N moves a box 3 m along the line of the force. Find the work.
3How many joules are in 3.5 kJ?
4Is work a vector or a scalar?
5Name the three conditions needed for a force to do work.
A force acts, the object has a displacement, and the force acts along the line of the displacement (at least partly).
More practice: on your own
1A snowplough pushes snow with a constant 4.5 × 10³ N force for 250 m. How much work does it do on the snow? Give the answer in megajoules.
W = (4.5 × 10³ N)(250 m) = 1.125 × 10⁶ J ≈ 1.1 MJ.
2A crane does 3.9 × 10⁴ J of work lifting a steel beam of mass 400 kg at constant velocity. How high does it lift the beam?
F = mg = (400)(9.8) = 3920 N. Δd = W ÷ F = 39 000 ÷ 3920 ≈ 9.9 m.
3A child throws a ball straight up. While the ball rises 3.0 m, does gravity do positive or negative work on it? If the ball's mass is 0.20 kg, how much?
Gravity pulls down while the ball moves up: negative work. W = −mgΔd = −(0.20)(9.8)(3.0) = −5.9 J.
4A satellite moves in a circular orbit at constant speed. Gravity always pulls toward Earth's centre, at right angles to the satellite's motion. How much work does gravity do on it? What does this tell you about its speed?
Gravity is always perpendicular to the motion, so it does zero work. With no work done, the satellite's kinetic energy, and so its speed, stays the same.
5Two students each push a 20 kg box. Ana pushes hers 5.0 m with 60 N; Ben pushes his 3.0 m with 90 N. Who does more work? Show your reasoning.
Ana: 60 × 5.0 = 300 J. Ben: 90 × 3.0 = 270 J. Ana does more work, even though Ben pushes harder.
Challenge
1A 15 kg crate is pushed 6.0 m across a level floor at constant velocity by a horizontal force. The coefficient of kinetic friction is 0.40. Find the work done by (a) the applied force, (b) friction, (c) gravity. Use g = 9.8 N/kg.
FN = mg = 147 N, so Ff = μFN = 0.40 × 147 = 58.8 N. At constant velocity, F(applied) = Ff = 58.8 N. (a) W = (58.8)(6.0) = +353 J ≈ +3.5 × 10² J. (b) W = −3.5 × 10² J. (c) Gravity is perpendicular to the motion: 0 J. (The total work is zero, which matches the constant speed.)
2Using the Going further tip: a student pulls a wagon 25 m with a handle tension of 80 N at 30° above the horizontal. How much work does the tension do?
W = FΔd cos θ = (80 N)(25 m)(cos 30°) = 2000 × 0.866 ≈ 1.7 × 10³ J. Only the horizontal part of the pull, about 69 N, does work.
3You lift a 5.0 kg bag of potatoes from the floor to a 1.0 m counter, then put it back on the floor. What total work do you do on the bag? What total work does gravity do? Explain.
Up: you do +49 J and gravity does −49 J. Down (lowering slowly): your upward force is opposite the downward motion, so you do −49 J, and gravity does +49 J. Totals: 0 J for you and 0 J for gravity. The bag ends where it started, so there is no net change in its energy.
Quiz
1. Which situation involves work being done on the object?
2. A 25 N force pushes a box 4.0 m along the line of the force. The work done is:
3. The unit joule is equivalent to:
4. Friction acts on a sliding puck. The work done by friction on the puck is:
5. A student does 450 J of work lifting a box 1.5 m at constant velocity. What is the box's weight?
6. Why does gravity do no work on a book sliding across a level table?
Big idea: Work is energy transferred by a force acting along a displacement (D3.5, D2.2). The joule links work and energy, which the rest of the unit builds on.
How to help: Act out the cases. Push on a wall (no work), lift a grocery bag (work), carry it across the kitchen (no work by the upward force). Ask your teen to name the force, the displacement and whether they are along the same line.
Common difficulties:
- Confusing effort with work. Tired muscles do not mean work was done on the object.
- Using mass instead of weight. To lift at constant velocity, the force is mg (in newtons), not m (in kilograms).
- Forgetting the negative sign for friction or for gravity on a rising object.
- Unit slips: kJ to J (× 1000), and cm to m (÷ 100).
The free-body diagrams, friction coefficients and g = 9.8 N/kg used here come from the Forces unit.