The shape of the lesson
By the end of the lesson, learners can
- Work is force along the motion: In the lab, keep the pulling force fixed and raise the angle of the rope from \(0^\circ\) towards \(90^\circ\).
- Against gravity and friction: Lifting a mass through a height \(h\) needs work \(mgh\) against gravity, whatever path it takes.
- Constant speed: work in = work out: At constant speed the resultant force is zero, so the pulling force does exactly as much work as is done against friction, resistance and gravity.
- Work \(= Fd\cos\theta\), \(\theta\) between force and motion.
- Perpendicular forces (normal reaction on a level floor) do no work.
- Gravity: \(mg \times\) vertical height.
- Friction: find \(R\) first, then \(\mu R \times d\).
- Constant speed: work done by the driving force \(=\) total work done against the resistances.
One picture every idea returns to
Work done by a constant force \(= Fd\cos\theta\), where \(d\) is the distance moved and \(\theta\) the angle between the force and the direction of motion. Work is measured in joules (\(1\ \text{J} = 1\ \text{N m}\)). Work against gravity \(= mgh\); work against friction \(= Fd\) with \(F = \mu R\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Force times distance, along the way.. A force does work when its point of application moves. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Work is force along the motion | In the lab, keep the pulling force fixed and raise the angle of the rope from \(0^\circ\) towards \(90^\circ\). Project tab 1 of the lab (Work, energy and power lab, work done and power) and let a student drive it while the class predicts. Model the worked example A rope at an angle (Fig. 1) on the board, one line at a time. Your turn: A force of 40 N acts at \(60^\circ\) to the direction of motion of a particle that moves 5 m. Find the work done by the force. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Against gravity and friction | Lifting a mass through a height \(h\) needs work \(mgh\) against gravity, whatever path it takes. Model the worked example Lifting at constant speed on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Constant speed: work in = work out | At constant speed the resultant force is zero, so the pulling force does exactly as much work as is done against friction, resistance and gravity. Model the worked example Dragged at an angle on a rough floor on the board, one line at a time. | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 21 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Constant speed: work in = work out, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Do not use the full force when it is at an angle: \(60 \times 8 = 480\) J is the work only if the rope is horizontal. | Return to Work is force along the motion: ask a pair to state the correct version and show it on the lab before moving on. |
| Work done by the weight on a slope uses the VERTICAL height, \(mg \times d\sin\alpha\), not \(mg \times d\). | The distance along the slope only multiplies the component \(mg\sin\alpha\). |
| With the rope at an angle, the normal reaction is NOT \(mg\). | Using \(\mu mg\) for friction in the first example gives 800 J instead of 650 J. |
Work done by a force: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
A box is pushed 4 m across a level floor by a horizontal force of 25 N. Find the work done by the push, and state the work done by the weight of the box.
A rope with tension 60 N inclined at \(25^\circ\) above the horizontal pulls a crate 8 m along a horizontal floor. Find the work done by the tension.
A crane lifts a load of mass 200 kg through 15 m at constant speed, then lowers it back to its starting point at constant speed. Find the work done by the crane cable on the way up, and the total work done by gravity over the whole trip.
A crate of mass 20 kg is pulled at constant speed through 10 m across rough horizontal ground by a rope inclined at \(30^\circ\) above the horizontal. The coefficient of friction is 0.4. Find the tension and the work done by the rope.
A particle of mass 3 kg slides 6 m down a line of greatest slope of a rough plane inclined at \(30^\circ\), \(\mu = 0.2\). Find (a) the work done by the weight, (b) the work done against friction, (c) the net work done on the particle.
A sledge of mass 15 kg is pulled 50 m up a line of greatest slope of a hill inclined at \(10^\circ\) to the horizontal, at constant speed, by a rope parallel to the slope. The coefficient of friction is 0.1. Find the work done by the rope.
Answer key and marking notes
- Push: \(25 \times 4 = 100\) J.
- Weight: 0 J, because it is perpendicular to the motion.
- \(W = 60 \times 8 \times \cos 25^\circ\).
- \(W = 435\) J.
- Up: \(T = 2000\) N, so work \(= 2000 \times 15 = 30\,000\) J.
- Gravity does \(-30\,000\) J going up and \(+30\,000\) J coming down.
- Total work by gravity \(= 0\): it depends only on the overall change in height.
- \(R = 200 - T\sin 30^\circ = 200 - 0.5T\).
- \(T\cos 30^\circ = 0.4(200 - 0.5T)\), so \(T(0.8660 + 0.2) = 80\).
- \(T = 75.0\) N.
- Work \(= 75.05 \times 10 \times \cos 30^\circ = 650\) J.
- (a) \(30 \times 6\sin 30^\circ = 90\) J.
- (b) \(0.2 \times 30\cos 30^\circ \times 6 = 31.2\) J.
- (c) \(90 - 31.18 = 58.8\) J (the normal reaction does no work).
- Work against gravity \(= 150 \times 50\sin 10^\circ = 1302\) J.
- \(R = 150\cos 10^\circ = 147.7\) N; work against friction \(= 0.1 \times 147.7 \times 50 = 738.6\) J.
- Constant speed: work by the rope \(= 1302 + 738.6 = 2040\) J (3 s.f.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A crate is pulled 12 m along a horizontal floor by a rope with tension 50 N inclined at \(35^\circ\) above the horizontal, as shown. Find the work done by the tension.
- A491 J
- B600 J
- C344 J
- D41.0 J
A block slides along a rough horizontal floor. Which force acting on the block does NO work?
- AThe normal reaction from the floor
- BThe friction force
- CA horizontal pushing force
- DAir resistance
A crane lifts a load of mass 150 kg vertically through 20 m at constant speed. Find the work done by the crane.
- A30 000 J
- B3000 J
- C1500 J
- D15 000 J
A particle of mass 4 kg slides 10 m down a line of greatest slope of a plane inclined at \(25^\circ\) to the horizontal. Find the work done by its weight.
- A169 J
- B400 J
- C363 J
- D16.9 J
A crate of mass 25 kg is pulled at constant speed for 15 m across a rough horizontal floor by a rope inclined at \(20^\circ\) above the horizontal. The coefficient of friction is 0.3. Find the work done by the tension.
- A1010 J
- B1130 J
- C1080 J
- D1200 J
A sledge of mass 20 kg is pulled 40 m up a slope inclined at \(15^\circ\) to the horizontal, at constant speed, by a rope parallel to the slope. The coefficient of friction is 0.12. Find the work done by the rope.
- A3000 J
- B2070 J
- C3030 J
- D1140 J
A particle of mass 2 kg slides 8 m down a line of greatest slope of a rough plane inclined at \(30^\circ\) to the horizontal. The coefficient of friction is 0.25. Find the work done against friction.
- A34.6 J
- B40 J
- C20 J
- D80 J
A box is pushed 5 m up a line of greatest slope of a plane inclined at \(20^\circ\) to the horizontal by a horizontal force of 60 N, as shown. Find the work done by the 60 N force.
- A282 J
- B300 J
- C103 J
- D319 J
Quiz key
Work \(= Fd\cos\theta = 50 \times 12 \times \cos 35^\circ = 491\) J. The tempting 600 J uses the full tension, which is only right if the rope is horizontal. 344 J uses \(\sin 35^\circ\), the component that does no work here, and 41.0 J is the horizontal component of the force (in newtons), not the work.
The normal reaction is vertical and the motion is horizontal, so \(\cos 90^\circ = 0\) and it does no work. The tempting "friction" does do work: it acts along the line of motion (backwards), so work is done AGAINST it. A horizontal push and air resistance also act along the line of motion.
The lifting force is \(mg = 1500\) N, so the work done is \(1500 \times 20 = 30\,000\) J (the work done against gravity, \(mgh\)). The tempting 3000 J forgets \(g\) (\(150 \times 20\)). 1500 is the weight in newtons, a force, not work; and 15 000 J brings in a \(\tfrac{1}{2}\) as if this were kinetic energy.
Vertical drop \(= 10\sin 25^\circ = 4.226\) m, so the work done by the weight is \(40 \times 4.226 = 169\) J (equivalently, the component \(40\sin 25^\circ\) times the 10 m). The tempting 400 J uses the distance along the slope with the full weight. 363 J uses \(\cos 25^\circ\), and 16.9 is the weight component in newtons, not the work.
Vertically: \(R = 250 - T\sin 20^\circ\). Horizontally: \(T\cos 20^\circ = 0.3R\). So \(T(\cos 20^\circ + 0.3\sin 20^\circ) = 75\), giving \(T = 71.96\) N. Work \(= T\cos 20^\circ \times 15 = 1010\) J (3 s.f.). The tempting 1130 J uses friction \(\mu mg = 75\) N, ignoring the lift from the rope. 1080 J is \(T \times 15\), without the \(\cos 20^\circ\), and 1200 J uses \(R = 250\) to get \(T = 75 \div \cos 20^\circ\), then multiplies \(T\) by 15 without the \(\cos 20^\circ\).
Against gravity: \(200\sin 15^\circ \times 40 = 2070.6\) J. Against friction: \(0.12 \times 200\cos 15^\circ \times 40 = 927.3\) J. Total \(= 2997.8 \approx 3000\) J. The tempting 2070 J leaves out friction. 3030 J uses friction \(\mu mg\) instead of \(\mu mg\cos 15^\circ\), and 1140 J subtracts the friction as if it helped the rope.
\(R = 20\cos 30^\circ = 17.32\) N, so friction \(= 0.25 \times 17.32 = 4.330\) N, and the work done against it is \(4.330 \times 8 = 34.6\) J. The tempting 40 J uses \(R = mg = 20\) N, true only on level ground. 20 J uses \(\sin 30^\circ\) for the normal reaction, and 80 J is the work done BY the weight (\(20\sin 30^\circ \times 8\)).
The motion is along the slope, at \(20^\circ\) to the horizontal force, so work \(= 60 \times 5 \times \cos 20^\circ = 282\) J. The tempting 300 J ignores the angle, but part of the push presses the box into the plane and does no work. 103 J uses \(\sin 20^\circ\) (the perpendicular component), and 319 J divides by \(\cos 20^\circ\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.