Welcome to Work & Energy. The Newton's Laws topic followed a body from moment to moment: the net force sets the acceleration, and the acceleration changes the velocity. This topic follows the same bodies with a different kind of bookkeeping. Instead of asking how quickly the velocity changes, it asks what a force has achieved once the body has moved some distance. That quantity is work, and it is the link between forces and energy. This page defines work and practises finding it: the work of one force, along the motion or at an angle to it; positive, negative and zero work; the work of friction; and the total work of all the forces on a body. The next page measures how quickly work is done, which is power. The third shows that the total work on a body equals the change in its kinetic energy, and the last three build potential energy, its conservation and the energy that friction turns into heat.
Work & Energy is the largest topic in the physics paper: 21 of the 192 questions so far, about five a paper. One kind of question has come up in every paper, always early, between Q18 and Q22: the work done by one named force. December 2025 Q18 and April 2026 Q19 are the same question with new numbers, a horizontal pull with friction also acting. January 2026 Q21 and March 2026 Q22 pull at an angle instead. December 2025 Q47 asks for the total work of all the forces, a few steps further on. All five appear below, as worked examples or as check-ins. Their wrong options come from a short list of slips: the work of the wrong force, the wrong trigonometric ratio, or a lost minus sign. This page names each one as it comes up.
What work measures
In everyday speech, holding a heavy suitcase still for ten minutes is hard work. In physics it is no work at all, because the suitcase does not move. Physics keeps the word for one thing: a force acting on a body while the body moves in the direction of the force. A crane lifting a load, a horse pulling a cart along a road and gravity pulling a falling stone all do work. A hand holding a suitcase still, or pushing on a wall that does not move, does none.
How much work? Think about the crane. Lifting a 1000 N load through 20 m is the same job as lifting it through 10 m twice, so it takes twice as much effort, twice the fuel. Lifting two such loads, 2000 N, through 10 m also takes twice as much. The amount grows in proportion to the force and in proportion to the distance moved along it, and the simplest quantity that does both is their product. For a constant force F that points along the body's displacement s, the work done by the force is
W = Fs.
The unit is the newton metre, called the joule: 1 J = 1 N·m, the work done when a force of 1 N moves a body 1 m along the force. A 20 N pull that drags a box 5 m along the pull does 20 × 5 = 100 J of work.
Work is a scalar. F and s are vectors, but the work is a plain number with no direction. A 10 N pull that moves a box 5 m along it does 50 J whether the box goes north, east or up a ramp. April 2026 Q1 asked which of displacement, acceleration, force and work is a scalar; the answer was work. A work can still be negative, as the sections below show, but its minus sign is not a direction. It says whether the force helps the motion or resists it.
Work. A constant force F acting along a body's displacement s does work W = Fs on the body. Work is a scalar, measured in joules: 1 J = 1 N·m. No displacement, no work.
Try it! · Check-in question
Which statement about work and its unit is correct?
W = Fs multiplies a force in newtons by a distance in metres, so 1 J = 1 N·m. Work has no direction: a 10 N pull over 5 m does 50 J whichever way the box goes, so work is a scalar, and the answer is B. A has the right unit, but work does not point anywhere; even a negative work carries a sign, not a direction. C, N/m, is the unit of a spring's stiffness k from the Elastic Force & Hooke's Law page, a force divided by a length. D, 1 kg·m/s², is the newton itself, the unit of force; the force is the vector here, not the work.
Whose work? One force at a time
Most bodies in the papers feel several forces at once: the weight, the normal force, a pull, friction. Each force does its own work, and a question about "the work done by F" asks about F alone. That work is F times the distance moved along F, and nothing else enters it. Friction does not shrink the pull's work: a 15 N pull is still 15 N whatever else acts on the box, and the box still moves 3 m along it. The friction does separate work of its own, which a later section works out. The mass does not enter either.
So read the end of the stem before you calculate. "The work done by the pulling force" is the work of one force. "The total work", or "the work done by the resultant force", is the work of all the forces together, the subject of the last section. The papers put both answers among the options.
Worked example · the work of one force, with friction acting (April 2026, Q19)
A horizontal pulling force of 15 N moves an object a distance of 3 m on a rough surface. If the frictional force is 5 N, the work done by the pulling force is ( )
A. 45 J B. 60 J C. 30 J D. 15 J
Step 1, whose work: the question names the pulling force, so only the pull counts.
Step 2, the direction: the pull is horizontal, and the object moves along the horizontal surface, the way the pull points. The pull acts along the displacement.
Step 3: W = Fs = 15 × 3 = 45 J. The answer is A.
C, 30 J, is (15 − 5) × 3, the total work of the pull and the friction together. That answers a different question, and April 2026 Q42, later in the same paper, asked it with new numbers. D, 15 J, is 5 × 3, the size of the friction's work. B, 60 J, is (15 + 5) × 3, which adds the friction to the pull, as if friction helped the object along.
Try it! · Check-in question
A horizontal force F = 20 N is applied to a 10 kg object, causing it to move 5 m along a rough horizontal surface with the friction force f = 5 N. The work done by F is ( ). (December 2025, Q18)
The question asks for the work done by F, so only F counts. F is horizontal and the object moves along the horizontal surface, so W = Fs = 20 × 5 = 100 J, and the answer is D. A, 75 J, is (20 − 5) × 5, the total work of F and the friction together. B, 25 J, is the size of the friction's work, 5 × 5, and C, −25 J, is the friction's work with its sign, as the section on friction shows. Both describe the friction, not F. The 10 kg mass is not needed either: on a level surface the weight is at right angles to the motion and does no work. April 2026 Q19 is this question with new numbers.
A force at an angle
Often the force does not point along the motion. A box pulled by a rope that slants upward still slides along the floor. How much work does the rope do?
Use the tool from the Adding & Resolving Forces page: replace the pull F by two parts at right angles. If θ is the angle between the rope and the floor, one part, F cos θ, lies along the floor, and the other, F sin θ, points straight up. The box moves along the floor, so the first part acts along the displacement and does (F cos θ) × s of work. The box never moves upward, so the upward part has no displacement along it and does no work; it only takes some of the weight off the floor. The work of the whole pull is therefore
W = Fs cos θ,
where θ is the angle between the force and the displacement. The same product can be read the other way round, as F times s cos θ, the part of the displacement that lies along the force. Both readings give the same number. The second is handy when the force is simple but the path slants, as in Practice Question 2.
Two checks. When θ = 0° the force lies along the motion, cos 0° = 1 and W = Fs, as in the first section. When θ = 90° the force is at right angles to the motion, cos 90° = 0 and W = 0.
The pull in January 2026 Q21, split into two parts at right angles. The box slides along the floor, so only the part along the floor, F cos 37° = 40 N, acts along the displacement and does work: 40 × 10 = 400 J. The upward part, F sin 37° = 30 N, has no displacement along it and does none. In general, W = Fs cos θ.
Work done by a constant force. W = Fs cos θ, where θ is the angle between the force and the displacement. Only the part of the force along the displacement, F cos θ, does work. The formula needs a constant force and a straight-line displacement.
Worked example · a pull at an angle (January 2026, Q21)
A wooden box is pulled by a force F = 50 N at an angle of 37° to the horizontal direction, which causes it to move a distance of 10 m on a horizontal surface. Given that cos 37° ≈ 0.8, the work done by the force F on the box is
A. 400 J B. 300 J C. 500 J D. 250 J
Step 1, the angle: the box moves along the horizontal surface, and F makes 37° with the horizontal, so the angle between the force and the displacement is θ = 37°.
Step 2: W = Fs cos θ = 50 × 10 × 0.8 = 400 J. The answer is A.
B, 300 J, uses sin 37° ≈ 0.6 in place of cos 37°. That is the part of the pull that points up, and the box never moves up. C, 500 J, ignores the angle, as if the whole pull lay along the floor. D, 250 J, is half of 500 J and does not follow from any common slip.
March 2026 Q22 asked the same question with letters in place of numbers. A symbolic question needs no new idea: find the angle between the force and the displacement, and the work is the force times the distance times its cosine.
Try it! · Check-in question
A block moves a distance x₀ horizontally towards the right under a force F at angle α to the horizontal. The work done by F is ( ) (March 2026, Q22, with its misprint corrected)
The block moves horizontally, and F makes the angle α with the horizontal, so the angle between the force and the displacement is α and W = Fx₀ cos α, so the answer is A. C uses sin α, which gives the part of F that points up; the block does not rise, so that part does no work. D divides by cos α, which makes the work larger than Fx₀, more than the same force would do pulling straight along the floor. The part of F along the motion, F cos α, is smaller than F, never larger. B divides by sin α, which would make the work grow without limit as the pull became nearly horizontal. Neither division comes from resolving the force.
Sin or cos? The part of the force along the displacement is the side of the force triangle next to the angle θ, so the work always takes cos θ, where θ is measured between the force and the displacement. If a question gives the angle β between the force and the vertical instead, the angle with a horizontal displacement is 90° − β, and the work is Fs cos(90° − β) = Fs sin β. Find the angle between the force and the displacement first, then take its cosine. A push that slants below the horizontal works the same way as a pull that slants above it.
Exam wording. In Chinese papers work is 功 and the joule is 焦耳 (J). "The work done by the pulling force" is 拉力做的功. Positive and negative work are 正功 and 负功, and doing work against friction is 克服摩擦力做功. The total work is 总功, and the work of the net external force is 合外力做的功. March 2026 Q22 has a misprint: its stem names the distance x₀, but options A, C and D write x. They mean the same distance, and A is the intended answer.
Positive, negative and zero work
W = Fs cos θ has a sign as well as a size, and the sign comes from cos θ. Between 0° and 90°, cos θ is positive: a force with a part along the motion does positive work. At 90°, cos θ = 0 and the work is zero. Between 90° and 180°, cos θ is negative: the force has a part pointing against the motion, and its work is negative. At 180° the force points straight back against the motion, cos 180° = −1 and W = −Fs.
The sign of a force's work comes from the angle θ between the force F and the displacement s (grey). Left: F has a part along the motion, cos θ > 0 and the work is positive. Middle: F is at right angles to the motion and does no work. Right: F has a part against the motion, cos θ < 0 and the work is negative.
Positive work means the force helps the motion along; negative work means it resists the motion. The Kinetic Energy page makes this exact: positive work on a body adds to its kinetic energy, and negative work takes kinetic energy away. A force that does −15 J of work on a body is often described the other way round: the body does 15 J of work against that force, or "overcomes" it. Both sentences describe the same 15 J.
Zero work comes about in two ways. Either the body does not move, like the suitcase held still or the wall pushed on, or the force is at right angles to the motion. On a level floor the weight points down and the normal force up, while the body moves sideways, so neither does any work, however large it is. A shopper carrying a bag along a level corridor holds it up with an upward force, but the bag moves horizontally, so that force does no work on the bag either.
The sign of work. 0° ≤ θ < 90°: W > 0, and the force helps the motion. θ = 90°: W = 0. 90° < θ ≤ 180°: W < 0, and the force resists the motion. A force on a body that does not move does no work.
Try it! · Check-in question
A ball is thrown straight up and later falls back into the thrower's hand. Ignoring air resistance, what is the sign of the work done by gravity on the ball while it rises, and while it falls?
Gravity points down all the time. While the ball rises, its displacement points up, opposite to gravity, so θ = 180° and gravity's work is negative: −mgh for a rise h. While it falls, the displacement points down, along gravity, so θ = 0° and the work is +mgh. The answer is C. A has the signs the wrong way round: on the way up gravity pulls against the motion, which is why the ball slows. B takes the sign from the force alone, but it depends on the angle between the force and the displacement, and the displacement reverses at the top. D confuses being the only force with doing no work; zero work needs no displacement or a right angle, and here gravity lies along the motion. Over the whole trip the two works cancel, −mgh + mgh = 0. January 2026 Q26 starts from the first half: gravity does −70 J of work on a body on its way up. The Potential Energy page finishes that question.
W = Fs cos θ needs a constant force on a straight path. When the force turns as the body moves, split the path into short pieces that are nearly straight, find the work on each piece and add the pieces up. A stone whirled round a circle on a string is the clearest case. The string pulls toward the centre and the stone moves along the tangent, so on every piece the pull is at 90° to the motion and does no work. The total is zero, however far the stone goes. This fits the Circular Motion & Gravitation topic: the centripetal force, at right angles to the velocity at every instant, turns the velocity without changing the speed. The Kinetic Energy page shows why a force that does no work cannot change a speed.
Try it! · Check-in question
A stone tied to a string 0.5 m long moves at a steady speed in a circle on a smooth, level table, and the tension in the string is 8 N. How much work does the tension do on the stone while the stone goes half a turn?
The tension points along the string, toward the centre, and the stone always moves along the tangent, at right angles to the string. The tension turns as the stone goes round, so W = Fs cos θ cannot be used for the half-turn in one go. Split the path into short pieces instead: on each one the tension is at 90° to the motion and does no work. Adding zeros gives zero, so the answer is A. D multiplies the tension by the arc length, π × 0.5 ≈ 1.57 m, as if the tension pointed along the path. B multiplies it by the half-turn's displacement, the diameter of 1 m, as if the tension were one constant force pointing across the circle. C multiplies it by the radius, but the stone never moves along the string at all.
You will meet this again. The words positive work and negative work come back in electric-field questions. December 2025 Q41 says that the electric force does negative work of 8 × 10⁻⁴ J on a charge moved from A to B, and April 2026 Q41 says that it does 1 × 10⁻³ J of positive work. They mean exactly what they mean here: in December's question the electric force resisted the motion, and in April's it helped.
The work done by friction
Sliding friction from a fixed surface always points against the sliding. While a body slides along a straight line, the friction is at 180° to the displacement, so cos θ = −1 and
Wf = −fs.
In April 2026 Q19 the 5 N friction does −5 × 3 = −15 J of work on the object while the pull does +45 J; option D was the size of the friction's work. When a question gives μ instead of f, find the normal force first, as on the Friction page, and then f = μN. On a level floor with a horizontal pull, N = mg. With a rope pulling upward at θ, the Equilibrium: Zero Net Force page showed that the rope carries part of the weight, so N = mg − F sin θ and the friction is smaller.
Friction differs from a steady pull or from gravity in one way: it turns round whenever the body does. Push a box 4 m across a floor and then back again, and friction points backward on both legs, so both legs do negative work. The total is −f × 8 m, not zero, even though the box ends where it started. For friction, the distance that counts is the whole length of the path. Compare the thrown ball above: gravity kept pointing down, so its work on the way up and its work on the way down cancelled. The Potential Energy page builds on that difference.
Work done by sliding friction. Wf = −fs, where f = μN and s is the length of the path slid. It is negative on every leg, so it never cancels on a round trip.
Friction is not always negative. Wf = −fs is for a body sliding over a surface that stays still. Static friction can do positive work. When a lorry speeds up, friction from its floor drags a crate forward with it, and the crate moves forward, so that friction acts along the crate's displacement and does positive work on it. As for every force, the sign comes from the angle between the force and the displacement. The papers so far have asked only about sliding friction on a fixed floor.
Try it! · Check-in question
A box is pushed 4 m east across a level floor and then pushed 4 m back west to where it started. The sliding friction on it is 10 N throughout. How much work does friction do on the box over the whole trip?
Friction always points against the sliding: west on the way out and east on the way back. On each leg it is at 180° to the displacement, so each leg gives −10 × 4 = −40 J, and the whole trip gives −40 + (−40) = −80 J. The answer is C. A puts the trip's overall displacement, zero, into W = Fs cos θ, but that formula needs a constant force, and friction reversed halfway. Split the trip where the force changes. B has the right size but the wrong sign: friction resists the motion on both legs. D counts only one leg.
The total work
When several forces act, each does its own work, and because work is a scalar, the total work is the plain sum of the separate works, signs included:
Wtotal = W₁ + W₂ + W₃ + …
Take April's object once more. The pull does +45 J and the friction −15 J, while the weight and the normal force, at right angles to the floor, do nothing. The total is 45 − 15 + 0 + 0 = 30 J.
There is a second route. The net force on the object is 15 − 5 = 10 N along the floor, and 10 × 3 = 30 J, the same answer. That is no accident. Each force's work is its part along the displacement times s, so adding the works adds up those parts and multiplies the sum by s. The parts along the displacement add up to the part of the net force along it. So when the forces are constant and the path is straight,
Wtotal = Fnet s cos θ,
with θ the angle between the net force and the displacement. Use whichever route is shorter. If the net force is known, or comes out of F = ma, the second is quicker. If the forces' works are easy one at a time, use the first.
April 2026 Q19 again, with every force's work. The pull does +45 J and the friction −15 J; the weight and the normal force, at right angles to the displacement, do none. The question asked for the pull's work alone, 45 J. The total work is 30 J, the same as the net force, 10 N, times 3 m. The pull and friction arrows are drawn to scale; N and mg are not.
One consequence is worth knowing. A body moving at a constant velocity has zero net force on it, by the first law, so the total work on it is zero, even though the pull and the friction may each do plenty of work. Their works cancel exactly. The Kinetic Energy page explains the pattern: the total work equals the change in kinetic energy, and at a constant velocity that change is zero.
Whose work is asked for? "The work done by F" means that force alone: Fs cos θ, whatever else acts. "The total work", "the net work" or "the work done by the resultant force" means the sum of every force's work, or Fnet s cos θ. April 2026 tested both: Q19 asked for the pull's work, and Q42 for the gain in kinetic energy, which comes from the total work.
Worked example · the total work, with the distance found first (December 2025, Q47)
As shown in the figure, a uniform electric field is present in the horizontal direction on a horizontal desktop. A block of mass m = 0.1 kg carrying a charge Q = −2.0 × 10⁻⁴ C is placed at point O on the desktop and released from rest. The sliding friction between the block and the desktop is f = 0.2 N, and the block experiences an electric field force of F = 0.4 N and is directed to the right. The net work done by the resultant external force in 2 s is ( ).
The figure shows the block at rest at point O on the desktop, in a uniform horizontal field E.
A. 3.2 J B. 1.6 J C. 0.8 J D. 2.4 J
Step 1, what matters: the electric force is given directly, 0.4 N to the right, so the charge and the field are not needed. A negative charge is pushed against the field, but the stem has already said which way the force points. The 0.4 N is more than the 0.2 N friction can supply, so the block slides to the right, and the friction points left.
Step 2, the net force and the distance: Fnet = 0.4 − 0.2 = 0.2 N, so a = Fnet/m = 0.2 ÷ 0.1 = 2 m/s². From rest, s = ½at² = ½ × 2 × 2² = 4 m.
Step 3, the total work: Wtotal = Fnets = 0.2 × 4 = 0.8 J. The answer is C.
Check by adding the works one at a time: the electric force does 0.4 × 4 = 1.6 J and the friction does −0.2 × 4 = −0.8 J, so the total is 1.6 − 0.8 = 0.8 J. The same 0.8 J is the block's kinetic energy after 2 s: v = at = 4 m/s, and ½ × 0.1 × 4² = 0.8 J. The Kinetic Energy page explains why the two agree.
B, 1.6 J, is the electric force's work alone, the answer to a question about F, not about the resultant. D, 2.4 J, adds the friction's 0.8 J instead of taking it away. A, 3.2 J, ignores friction completely: a = 0.4 ÷ 0.1 = 4 m/s², s = ½ × 4 × 2² = 8 m and 0.4 × 8 = 3.2 J.
Try it! · Check-in question
A box is pulled 5 m across a level floor by a rope at 37° above the horizontal. The tension in the rope is 50 N, and the sliding friction on the box is 20 N. What is the total work done on the box by all the forces? (cos 37° = 0.8)
Go force by force. The tension does 50 × 5 × cos 37° = 50 × 5 × 0.8 = 200 J. The friction does −20 × 5 = −100 J. The weight and the normal force are at right angles to the floor and do none. The total is 200 − 100 = 100 J, so the answer is B. The net-force route agrees: along the floor the net force is 40 − 20 = 20 N, and 20 × 5 = 100 J. A is the tension's work alone, the answer to a question about the rope. C ignores the angle: 50 × 5 − 100 = 150 J. D adds the friction's 100 J instead of taking it away; friction's work is negative.
The slips the options are built from. The work of the wrong force: the total work given for one force's work (December's 75 J, April's 30 J), or one force's work given for the total (December 2025 Q47's 1.6 J). The wrong ratio: sin θ in place of cos θ (January's 300 J), or the angle ignored (500 J). A lost sign: sliding friction's work is negative, and so is the work of any force with a part against the motion. Adding where you should subtract: friction resists, so its work comes off the total (April's 60 J, December 2025 Q47's 2.4 J).
Your turn! · Practice questions
Practice · Question 1
A gardener pushes a lawnmower 15 m across a level lawn. She pushes along the handle with a steady force of 40 N, directed 60° below the horizontal. How much work does her push do on the mower? (cos 60° = 0.5, sin 60° ≈ 0.87)
The mower moves horizontally, and the push makes 60° with the horizontal, so the angle between the force and the displacement is 60°, whether the push slants above the horizontal or below it. W = 40 × 15 × cos 60° = 40 × 15 × 0.5 = 300 J, so the answer is D. A, 600 J, ignores the angle, as if the whole push lay along the lawn. B uses sin 60°, which gives the part of the push that presses down into the lawn; the mower does not move into the ground, so that part does no work. C makes the work negative because the push points downward, but the sign depends on the angle between the force and the displacement, here 60°, less than 90°. The push has a part along the motion, so it helps, and its work is positive.
Practice · Question 2
A student carries a 50 N bag up a staircase, holding it with a steady upward force of 50 N. Over the climb the bag rises 3 m and moves 4 m horizontally, so its displacement is 5 m along the slope of the stairs. How much work does the student's upward force do on the bag?
The force points straight up, and of the bag's 5 m displacement, 3 m is upward. Read W = Fs cos θ as the force times the part of the displacement along it: 50 × 3 = 150 J. With the angle instead: between the vertical and the slope, cos θ = 3/5, so W = 50 × 5 × 3/5 = 150 J. The answer is B. A uses the whole 5 m, as if the bag moved straight up. C uses the 4 m across, the part of the displacement at right angles to the force, which takes no part in the work. D would be right for a bag held still, or carried along a level corridor, but here the bag rises 3 m along the force.
Practice · Question 3
A crane lowers a 500 kg crate 4 m at a constant speed. How much work does the tension in the crane's cable do on the crate? (g = 10 m/s²)
The crate moves at a constant speed in a straight line, so the net force on it is zero and the tension balances the weight: T = mg = 500 × 10 = 5000 N, upward. The crate moves 4 m downward, at 180° to the tension, so W = 5000 × 4 × cos 180° = −2.0 × 10⁴ J, and the answer is A. B has the right size but the wrong sign: the cable pulls up while the crate goes down, so the cable resists the motion. C confuses the tension's work with the total work. The total is zero, because gravity does +2.0 × 10⁴ J, but the tension's own work is not. D takes the tension as 500 N, the mass in kilograms instead of the weight in newtons.
Practice · Question 4
A 10 kg sledge is pulled 10 m across level snow by a rope at 37° above the horizontal. The tension in the rope is 50 N, and the coefficient of kinetic friction between the sledge and the snow is 0.2. How much work does friction do on the sledge? (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8)
Find the normal force first. The rope's upward part is 50 × sin 37° = 30 N, so N = mg − 30 = 100 − 30 = 70 N, and the friction is f = μN = 0.2 × 70 = 14 N. The sledge slides 10 m with the friction pointing backward, so Wf = −14 × 10 = −140 J, and the answer is D. A takes N = mg = 100 N, which gives f = 20 N and −200 J; the rope carries part of the weight, so the snow presses up less. B swaps sin and cos, taking the upward part as 50 × 0.8 = 40 N, so N = 60 N, f = 12 N and −120 J. C has the right size but the wrong sign: sliding friction points against the motion, so its work is negative.
Practice · Question 5
A 2 kg block is at rest on a smooth, level floor. A rope pulls it with a constant force of 10 N at 37° above the horizontal. What is the total work done on the block in the first 2 s? (sin 37° = 0.6, cos 37° = 0.8)
The floor is smooth, and the weight and the normal force are at right angles to the motion, so only the rope does work. Along the floor the net force is the rope's part F cos 37° = 10 × 0.8 = 8 N, so a = 8 ÷ 2 = 4 m/s², and from rest s = ½at² = ½ × 4 × 2² = 8 m. The total work is 8 N × 8 m = 64 J, or, force by force, 10 × 8 × 0.8 = 64 J. The answer is C. A ignores the angle twice: a = 10 ÷ 2 = 5 m/s², s = 10 m and 10 × 10 = 100 J. D uses the angle in the work but forgets it in the acceleration: s = 10 m and 10 × 10 × 0.8 = 80 J. B uses sin 37° throughout: a 6 N part along the floor gives a = 3 m/s², s = 6 m and 6 × 6 = 36 J.
Things to remember
Concept
What to know
Work
W = Fs for a constant force along the displacement. No displacement, no work. Work is a scalar (April 2026 Q1); 1 J = 1 N·m.
A force at an angle
W = Fs cos θ, with θ between the force and the displacement. Only F cos θ does work. January 2026 Q21: 50 × 10 × 0.8 = 400 J. March 2026 Q22: Fx₀ cos α.
Whose work?
"The work done by F" uses F alone; friction and the mass do not enter. December 2025 Q18: 20 × 5 = 100 J. April 2026 Q19: 15 × 3 = 45 J.
The sign
θ < 90°: positive, the force helps. θ = 90°: zero. θ > 90°: negative, the force resists. −15 J of work by a force means the body does 15 J of work against it.
Zero work
No displacement, or a force at right angles to the motion: the weight and the normal force on a level floor, a hand carrying a bag along a level corridor, the string on a body going round a circle.
Friction
Sliding friction on a fixed surface does Wf = −fs, with f = μN and s the length of the path. Find N first: with a rope pulling up at θ, N = mg − F sin θ. It never cancels on a round trip.
Total work
The sum of every force's work, signs included, or Fnet s cos θ. Zero at a constant velocity. December 2025 Q47: 0.2 N × 4 m = 0.8 J.
The planted slips
The total work given for one force's work, or the other way round; sin θ for cos θ; the angle ignored; friction's work added instead of subtracted; a lost minus sign.
