Work, Energy, and Simple Machines
Why This Matters
Think about your day. Food gives your body the energy to walk to school. Electricity gives a fan the energy to spin. Petrol gives a car the energy to move. The same single idea — energy — is hiding behind all of them.
In earlier chapters you learnt how forces change motion, and how Newton’s laws describe that motion. Those laws work well. But sometimes the force keeps changing, or acts in a messy way, and using the laws directly gets very hard. There is a smarter shortcut. Instead of tracking every push at every instant, we track how much energy went in. This often gives the answer in one or two clean steps.
This chapter builds that shortcut from scratch. You will learn what “work” really means in science — and why it is not the same as the everyday word. You will meet energy, in its many forms, and learn exactly how much energy a moving thing or a raised thing carries. You will see one of the most powerful rules in all of physics: energy is never made or destroyed, only changed from one form to another. And you will learn how a simple ramp, lever or pulley lets you lift things that feel far too heavy — and the clever catch that makes this possible.
The Big Idea
The Big Idea: Work is done when a force moves an object — and it equals force × distance moved in the direction of the force (W = F × s). Doing work on an object gives it energy, the capacity to do work back. Energy can be hidden in motion (kinetic) or in position (potential), and it can switch forms — but the total amount never changes. A simple machine cannot reduce the total work; it only lets you swap a big force over a short distance for a small force over a long distance.
Hold on to one sentence: work transfers energy, and energy is never lost — only moved around or changed in form. Almost every idea in this chapter is just that one sentence, made exact and then applied. Let us build it piece by piece.
What “Work” Means in Science
In everyday talk, “work” means anything tiring — studying, standing in a queue, holding a heavy bag. In science, “work” has a sharp, exact meaning. Let us build it from a simple experience: lifting bags.
Imagine a wheat bag of mass 5 kg on the floor. Gravity pulls it down with a force mg. To lift it slowly to a height of 1 metre, you must push up with a force equal to mg. The bag moves 1 m in the direction of your push. In everyday language, you “did some work”. Now watch how the amount changes.
- Lift 3 such bags one after another to the same height → you did 3 times the work.
- Lift all 3 bags together → you need 3 times the force, over the same 1 m → again 3 times the work.
- Lift 1 bag but to 3 metres instead of 1 m → 3 times the distance → again 3 times the work.
So the work done grows when the force grows, and grows when the distance grows. That is exactly what the scientific definition captures.
Work done by a constant force = force applied × displacement in the direction of the force
W = F × s
Figure 7.1 shows this for a horizontal push and for a straight lift. The arrow you care about is the force (red), and the distance you care about is the part of the motion along that force (green).
The unit of work — the joule
The SI unit of force is the newton (N) and the SI unit of distance is the metre (m). So the unit of work is newton × metre, which we call the joule (J), after the scientist James Prescott Joule.
1 joule = 1 newton × 1 metre
That is, 1 J of work is done when a force of 1 N moves an object 1 m in the direction of the force. Since 1 N = 1 kg m s⁻², we can also write 1 J = 1 kg m² s⁻² — a useful form when you check units in a problem.
Let us put the definition to work on a simple lift.
A boy lifts a 2 kg book from the floor onto a shelf 1.5 m high, slowly and steadily. How much work does he do on the book? Take g = 10 m s⁻².
- To lift the book slowly, the boy must push up with a force equal to the book’s weight. Weight = mg = 2 kg × 10 m s⁻² = 20 N. So the force F = 20 N, upward.
- The book moves up by s = 1.5 m, in the same direction as the force. So we use W = F × s directly.
- W = F × s = 20 N × 1.5 m = 30 J. The boy does 30 joules of work on the book.
When is the work zero?
Look hard at W = F × s. If either part is zero, the whole thing is zero. This leads to three surprising cases, shown in Figure 7.2.
Case 1 — no displacement (s = 0). Push hard against a rigid wall. You strain, you sweat, but the wall does not move. Since s = 0, the work done on the wall is zero. This feels wrong, because you feel tired! Here is why you tire even though no work is done. To keep pushing, the muscles in your body keep tightening and loosening. That uses up energy stored inside your body. So your body spends energy, but none of it is transferred to the wall as work. Tiredness is about your muscles, not about work done on the object.
Case 2 — no force (F = 0). If no force acts, no work is done, however far the object travels.
Case 3 — force perpendicular to the motion. Carry a box flat in your hands and walk across a room. Your hands push the box upward (to hold its weight), but the box moves sideways. The force is at right angles to the motion. There is no movement along the force, so the work done by your hands on the box is zero.
A waiter carries a tray of food at constant height while walking straight across a hall. How much work does the upward force from his hand do on the tray?
Zero. The hand’s force on the tray points up, but the tray moves horizontally. The force is perpendicular to the displacement, so there is no movement along the force, and the work done is zero.
Positive and negative work
Work can carry a sign. It is positive when the force and the displacement point the same way, and negative when they point opposite ways. Figure 7.3 shows both.
- Positive: Push a wheelchair forward. Your force and the wheelchair’s motion point the same way. You do positive work on it.
- Negative: A goalkeeper stops a moving ball. Her force pushes the ball backward, but the ball is still moving forward as it slows. Force and displacement are opposite, so she does negative work on the ball.
Note one subtle thing: force and displacement have directions, but work does not. Work is just a number with a + or − sign — not an arrow.
Let us see the negative sign appear in a real calculation.
While saving a goal, a goalkeeper's hands move back by 15 cm as she stops a ball, applying a force of 200 N against the ball's motion. How much work does she do on the ball?
- The goalkeeper’s force points opposite to the ball’s motion. So this is negative work. We take the displacement of the ball, in the direction of the applied force, as negative: s = −15 cm = −0.15 m.
- Work done = force × displacement in the direction of the force = 200 N × (−0.15 m).
- W = 200 N × (−0.15 m) = −30 J. The goalkeeper does −30 J of work on the ball. The minus sign tells us she is taking energy away from the ball (slowing it down).
Work and Energy Go Together
Throw a cricket ball at the stumps. The moving ball knocks the bails off. Raise a flowerpot high and let it fall — it can crack the floor below. In each case the ball or the pot was able to do work on something else. Anything that has the capacity to do work is said to possess energy.
But where did that energy come from? The ball got it from the work the fielder did when throwing it. The pot got it from the work done in lifting it up high. When you do positive work on an object, it gains energy. Later it can spend that energy to push on something else. This tight link is captured by a neat rule:
Work-energy idea: the work done on an object equals the change in its energy.
So work is not separate from energy — work is energy being transferred. The unit of energy is therefore the same as the unit of work: the joule (J).
Doing mechanical work is one way to transfer energy, but not the only way. Energy can also flow as heat (from a hot object to a cold one), as light and radiation (this is how the Sun’s energy reaches Earth across empty space), through electric circuits, and as sound.
A bowler throws a ball. Did the bowler do positive or negative work on the ball, and what happened to the ball's energy?
Positive work. The bowler’s force on the ball points the same way the ball moves, so the work is positive. By the work-energy idea, positive work means the ball gains energy — its energy increases as it speeds up.
The Many Forms of Energy
Energy is the capacity to do work, but it wears many different costumes. The examples so far were all mechanical energy (linked to motion and position). But energy also appears as heat, light, sound, electrical, chemical and nuclear energy — and it can switch from one form to another. Figure 7.4 lays them out.
You see these changes every day. A bulb turns electrical energy into light. A water heater turns electrical energy into thermal energy. The chemical energy in food powers your muscles and becomes mechanical energy. A ringing bell turns mechanical energy into sound. Among all of these, we will study mechanical energy most closely, because it connects directly to the forces and motion you already know.
Mechanical Energy — Kinetic and Potential
Mechanical energy is the energy an object has because of its motion or its position. It comes in two flavours: kinetic energy (motion) and potential energy (position). Let us pin down exactly how much of each an object has.
Kinetic energy — and where ½mv² comes from
The energy an object has because it is moving is called kinetic energy. Every moving thing has it — a rolling ball, a moving bus, a flying bird. A still object has zero kinetic energy.
How much kinetic energy does a moving object carry? Here is the beautiful part — we can derive it, not just memorise it. Imagine an object starting from rest. A constant force F pushes it through a distance s, speeding it up to a final velocity v (Figure 7.5). By the work-energy idea, the work done by F equals the energy the object gained — which is its kinetic energy.
Let us do the algebra. We need a general expression first, for an object of mass m speeding up from an initial velocity u to a final velocity v over a displacement s, under a constant force F.
Now follow the derivation step by step.
Derive the kinetic energy of an object of mass m moving with velocity v, starting from rest.
- Use the kinematic equation v² = u² + 2as and make s the subject: s = (v² − u²) ÷ (2a).
- The work done by the force is W = F × s. By Newton’s second law, F = ma. So W = ma × s.
- Substitute s from Step 1: W = ma × (v² − u²) ÷ (2a). The acceleration a cancels: W = ½ m (v² − u²).
- By the work-energy idea, this work equals the change in the object’s energy. So the change in energy = ½ m (v² − u²).
- If the object started from rest, u = 0, so the change in energy equals the final kinetic energy. Setting u = 0: kinetic energy K = ½mv². That is the formula, derived — not guessed.
Kinetic energy: K = ½mv²
Its unit is the joule. Kinetic energy has no direction — it is just a number. Speed the object up (positive work) and its kinetic energy rises. Slow it down (negative work) and its kinetic energy falls.
One thing trips up many students: kinetic energy depends on v², not v. So if speed doubles, kinetic energy goes up by 2² = 4 times — not 2 times. Let us check.
A car's speed doubles. What happens to its kinetic energy compared with the original?
- Let the mass be m and the original speed be v. Original kinetic energy = ½mv².
- The new speed is 2v. New kinetic energy = ½ m (2v)² = ½ m × 4v² = 4 × (½mv²).
- The new kinetic energy is 4 times the original. This is exactly why a car going twice as fast is far more dangerous — it carries four times the energy to get rid of when it stops.
An Indian fast bowler delivers a 0.2 kg cricket ball at about 43 m s⁻¹ (around 155 km h⁻¹). Find the kinetic energy of the ball.
- Use K = ½mv², with m = 0.2 kg and v = 43 m s⁻¹.
- K = ½ × 0.2 kg × (43 m s⁻¹)² = ½ × 0.2 × 1849 = 0.1 × 1849.
- K ≈ 184.9 J. The fast ball carries about 185 joules of kinetic energy — which is why it stings to catch.
Potential energy — energy stored by position or shape
The other kind of mechanical energy is potential energy — energy stored up, ready to be released. You meet it whenever you stretch, bend, lift or pull something apart.
Pull back a gulel (slingshot) and let go — the stone shoots off. Pull an archer’s bow and release — the arrow flies. Squeeze a spring and let it go — it pushes a block away. In each case you did work to deform the object (stretch the band, bend the bow, squeeze the spring). That work got stored in the object. When released, the object springs back to its shape and gives the stored energy to whatever it touches, as kinetic energy. Figure 7.6 shows two ways energy gets stored.
Energy can also be stored by changing the arrangement of objects, not just by bending one object. Two unlike magnetic poles attract; pull them apart and you store energy — release them and they rush together. Likewise a raised ball and the Earth attract each other through gravity; lift the ball and you store energy in the ball–Earth system. In general: whenever objects interact through a force — gravity, electric, magnetic — separating them against that force stores potential energy.
Gravitational potential energy: U = mgh
The simplest and most useful case is a ball lifted above the ground. The Earth is so much heavier than the ball that it barely moves, so we just call the stored energy the gravitational potential energy of the ball.
We can find a formula using the work-energy idea. Take an object of mass m on the ground, and call its potential energy zero there. To raise it slowly to a height h, you push up with a force equal to its weight, mg. The work you do is:
W = force × displacement = mg × h = mgh
By the work-energy idea, this work is stored as the object’s potential energy. So:
Gravitational potential energy: U = mgh
This says exactly what experience tells you: the higher you lift something (bigger h), and the heavier it is (bigger m), the more energy it stores. Drop a heavy ball into sand from a greater height and it digs a deeper pit — because it carried more energy. Its unit is, again, the joule.
Note: U = mgh is only accurate near the Earth’s surface, where g is roughly constant. Very far from Earth, g gets weaker and this simple formula no longer holds. You will study that in higher classes.
A fielder throws a 200 g cricket ball straight up so it reaches 10 m above the ground. How much potential energy does it have at the top? Take g = 10 m s⁻².
- Convert the mass to kilograms: 200 g = 0.2 kg. The height is h = 10 m.
- Use U = mgh = 0.2 kg × 10 m s⁻² × 10 m.
- U = 20 J. At the highest point the ball stores 20 joules of gravitational potential energy.
You carry a bag at the same height while walking forward across flat ground. Does its gravitational potential energy change?
No. Gravitational potential energy depends on height h. Walking forward at the same height does not change h, so the potential energy stays the same. It would only change if you raised or lowered the bag.
Conservation of Mechanical Energy
Here is where it all comes together. The mechanical energy of an object is the sum of its kinetic and potential energy. Now watch what happens to that sum as an object falls freely.
Lift an object to point A at height h and drop it (Figure 7.7). As it falls, its height drops — so its potential energy falls. But it speeds up — so its kinetic energy rises. The remarkable thing is that the amount lost from one exactly equals the amount gained by the other. The total stays fixed.
Let us prove it. At the start (point A), velocity is zero:
potential energy = mgh
kinetic energy = 0
mechanical energy = mgh + 0 = mgh
After falling for a time t, the object has dropped, and (using the kinematic equations with a = g) its speed is v = gt and it has fallen a distance ½gt². You can show that at this lower point:
potential energy = mgh − ½mg²t²
kinetic energy = ½mv² = ½mg²t²
mechanical energy = (mgh − ½mg²t²) + ½mg²t² = mgh
The two ½mg²t² terms cancel. The kinetic energy gained (½mg²t²) is exactly the potential energy lost (½mg²t²). So:
As an object falls under gravity alone, its mechanical energy stays constant. This is the conservation of mechanical energy.
A swinging pendulum shows the same thing. At the top of its swing it is momentarily still — all potential energy, no kinetic. At the bottom it moves fastest — all kinetic, no potential. It rises again to almost the same height it started from, because the total energy is conserved. (In real life it slowly stops, because friction and air resistance quietly drain a little energy as heat each swing — but the total energy, counting that heat, is still conserved.)
This conservation is a powerful shortcut. Newton’s laws can get messy; tracking the total energy often gets the final speed in one line.
A child slides down a frictionless slide of height h. Using energy conservation, find the speed at the bottom. Does it depend on the child's mass?
- At the top, the child is at rest: all energy is potential, PE = mgh, KE = 0.
- At the bottom, the height is zero: all energy is kinetic, KE = ½mv², PE = 0.
- Mechanical energy is conserved (no friction), so the energy at the top equals the energy at the bottom: mgh = ½mv².
- The mass m appears on both sides, so it cancels: gh = ½v², giving v² = 2gh, so v = √(2gh).
- The speed v = √(2gh) depends only on the height h. It does not depend on the child’s mass or the shape of the slide. So a heavy child and a light child reach the bottom of the same slide at the same speed.
A 10000 kg truck moving at 20 m s⁻¹ (72 km h⁻¹) has its brakes fail. It is steered onto a sandy escape ramp inclined at 30°, where sand pushes back with a force of 50000 N. What minimum ramp length stops the truck? (For a 30° ramp, the truck rises 1 m for every 2 m along the ramp.) Take g = 10 m s⁻².
- Find the truck’s starting kinetic energy: K = ½mv² = ½ × 10000 kg × (20 m s⁻¹)² = ½ × 10000 × 400 = 2000000 J. Its starting potential energy is 0, so total energy at the start = 2000000 J.
- Let the truck travel a distance d along the ramp before stopping. Using the hint, it rises a height h = d ÷ 2.
- At the stop, kinetic energy = 0, and potential energy gained = mgh = 10000 × 10 × (d ÷ 2) = 50000 × d. So the final total energy = 50000 × d (all of it potential).
- The sand does negative work, removing energy: work by sand = −50000 N × d. By the work-energy idea, this equals the change in the truck’s total energy: −50000 d = (50000 d) − 2000000.
- Rearrange: 2000000 = 50000 d + 50000 d = 100000 d. So d = 2000000 ÷ 100000 = 20 m. The ramp must be at least 20 m long to stop the truck.
Power — How Fast Work Is Done
Carry your school bag up to the classroom. Running up in one minute feels very different from strolling up in five minutes — even though the same work (mgh) is done both times. What differs is how fast you did it. That is power.
Power is the rate of doing work: P = W ÷ t (work done divided by the time taken).
To do more work in the same time, or the same work in less time, you need more power. The SI unit of power is the watt (W), named after James Watt: 1 watt = 1 joule per second (1 W = 1 J s⁻¹). Figure 7.8 shows the idea.
A weightlifter lifts a 75 kg mass by 2 m in 5 seconds. What power does she use? Take g = 10 m s⁻².
- First find the work done. Lifting against gravity: W = mgh = 75 kg × 10 m s⁻² × 2 m = 1500 J.
- Power is work divided by time: P = W ÷ t = 1500 J ÷ 5 s.
- P = 300 W. The weightlifter works at a rate of 300 watts during the lift.
Threads of curiosity — horsepower. You may have heard a car engine described in “horsepower (hp)”. One horsepower equals about 746 W. In the early days of engines, their power was compared to the power of real horses that used to pull carriages — and the name stuck.
Simple Machines
Often we need to do work against gravity or friction — lift a heavy box, raise a flag, pull out a stuck nail. The total work for a job cannot be reduced. But the job can be made easier by changing the size or the direction of the force you must apply. Devices that do this are called simple machines.
Two words you need: the effort is the force you apply, and the load is the force you must overcome (usually the weight of the thing you are moving). To measure how much a machine helps, we define its mechanical advantage:
Mechanical advantage = load ÷ effort
A mechanical advantage greater than 1 means the machine lets you overcome a big load with a smaller effort. We will look at three: the pulley, the inclined plane and the lever.
The pulley
A pulley is a wheel with a groove that guides a rope (Figure 7.9). Think of raising a flag: you pull the rope down, and the flag goes up.
A fixed pulley (the wheel stays in one place) does not reduce the force you need — it only changes its direction. Pulling down is much more comfortable than reaching up and lifting directly, because you can use your body weight. But the effort still equals the load, so its mechanical advantage is load ÷ effort = 1.
A movable pulley (or a system of pulleys) is different. Here the load hangs from a pulley that moves, and the rope’s tension supports the load from two sides. This can give a mechanical advantage greater than 1 — a small effort lifts a much heavier load. That is why cranes and lifts use pulley systems.
The inclined plane (ramp)
Suppose you must lift a heavy box onto a platform. Lifting it straight up (Figure 7.10a) needs a force equal to its full weight — maybe too much for you. Instead, lay a long smooth plank against the platform and push the box up the slope (Figure 7.10b). This needs a smaller force. But there is a catch: you must push over a longer distance.
Let us find the mechanical advantage. The load is the box’s weight, mg. Let the effort along the ramp be F′ and the ramp length be L, raising the box to height h. If you move it up at steady speed, the work you do (F′ × L) equals the potential energy it gains (mgh), ignoring friction:
F′ × L = mgh
So F′ = mg × (h ÷ L)
Now the mechanical advantage is load ÷ effort = mg ÷ F′:
Mechanical advantage of inclined plane = mg ÷ F′ = L ÷ h
Since the ramp length L is bigger than the height h, the effort F′ is less than the full weight mg, and the mechanical advantage is greater than 1. Make the ramp longer and gentler, and L ÷ h grows — so the effort needed shrinks. This is exactly why mountain roads wind gently instead of going straight up.
A person uses a ramp to lift an object over a step 30 cm high. The ramp has a horizontal width of 40 cm. What is its mechanical advantage?
- The ramp, the height and the width form a right-angled triangle. The height is 30 cm and the base is 40 cm; the ramp itself is the slanted side (the length L).
- Find L using the right-angle property: L² = 30² + 40² = 900 + 1600 = 2500, so L = √2500 = 50 cm. (This is the well-known 3-4-5 triangle.)
- Mechanical advantage = L ÷ h = 50 cm ÷ 30 cm = 1.67. The ramp lets the person lift the object using about 1.67 times less force than lifting it straight up.
The lever
A lever is a rigid bar that can turn about a fixed point. Balance a scale on a pencil, put a heavy stapler near the pencil on one end, and a light eraser far out on the other end — and the light eraser can lift the heavy stapler! How?
A lever has three parts (Figure 7.11): the fulcrum (the fixed turning point), the load (the force to overcome), and the effort (the force you apply). The distance from the load to the fulcrum is the load arm; the distance from the effort to the fulcrum is the effort arm.
Why does this work? When you push the effort end down by a large distance d₁, the load end rises by a small distance d₂. The work done on one end is passed to the other end, so:
F₁ × d₁ = F₂ × d₂
(effort × effort arm = load × load arm)
Rearranging, the force on the load is F₂ = F₁ × (d₁ ÷ d₂). If the effort arm d₁ is long and the load arm d₂ is short, a small effort F₁ produces a large force F₂ on the load. The mechanical advantage is:
Mechanical advantage of lever = load ÷ effort = effort arm ÷ load arm = d₁ ÷ d₂
So a longer effort arm means a bigger mechanical advantage. The catch is the same as always: the effort moves a longer distance, so the total work stays the same.
A seesaw has a fulcrum at C. A child of mass 15 kg sits 2 m from C on one side. Where should a 30 kg child sit on the other side to balance it?
- For balance, effort × effort arm = load × load arm: (15 kg)‘s weight × its distance = (30 kg)‘s weight × its distance. Since weight is mass × g and g is the same on both sides, g cancels, so we can just use masses: 15 kg × 2 m = 30 kg × L.
- Solve for L: L = (15 × 2) ÷ 30 = 30 ÷ 30 = 1 m.
- The 30 kg child should sit 1 m from the fulcrum. The heavier child sits closer in — that is why a lighter child far out can balance a heavier one near the centre.
Levers come in three classes, depending on which part — fulcrum, load or effort — sits in the middle.
| Class | What is in the middle | Everyday examples |
|---|---|---|
| Class I | Fulcrum in the middle (load and effort on either side) | Scissors, crowbar, pliers, seesaw, balance scale |
| Class II | Load in the middle | Wheelbarrow, bottle opener, lemon squeezer |
| Class III | Effort in the middle | Tweezers, tongs, broom, fishing rod |
A final big idea ties the whole chapter together: machines never create energy. In every machine, the work you put in equals the useful work done on the load (ignoring friction). A machine only helps you use energy more conveniently — it cannot give you energy for free. This is why a perpetual motion machine — one that runs forever doing useful work with no fuel — can never be built. Real machines always lose a little energy to friction as heat, so they slow down and stop unless you keep feeding them energy.
Common Mistakes
These are the trip-ups that catch students every year. Read each one slowly.
If you push really hard on something, you are doing work on it — even if it doesn't move.
In everyday speech, 'work' means anything that takes effort and makes you tired, so a hard push feels like obvious work.
Scientific work needs displacement. W = F × s, and if s = 0 (the object does not move), the work is zero no matter how hard you push. You feel tired because your muscles use energy, but no work is transferred to the object.
If an object's speed doubles, its kinetic energy also doubles.
We are used to things scaling in proportion, so it feels natural that twice the speed should mean twice the energy.
Kinetic energy is ½mv², which depends on v squared, not v. Double the speed and the energy goes up by 2² = 4 times. A car at twice the speed carries four times the kinetic energy.
A simple machine like a ramp or lever reduces the total work you have to do.
The machine clearly makes the job feel easier — you push with a smaller force — so it seems like the whole task got smaller.
The total work stays exactly the same. The machine only reduces the force, and in exchange you must move that force through a longer distance. Force × distance (the work) is unchanged; only the comfort changes.
A heavier object slides down a frictionless slide faster than a lighter one.
In daily life heavier things often feel more powerful and 'unstoppable', so we expect extra mass to mean extra speed.
From mgh = ½mv², the mass cancels, giving v = √(2gh). The speed at the bottom depends only on the height, not the mass. A heavy and a light object reach the bottom of the same frictionless slide at the same speed.
Doing the same job slower or faster changes the amount of work done.
Rushing a job feels harder and more demanding, so it seems like more work must be involved.
Work depends only on force and distance (W = F × s), not on time. Walking or running up the same stairs does the same work mgh. What changes is the power — the rate of doing work — not the work itself.
Quick Check
Test yourself before moving to practice.
A porter holds a heavy suitcase steady on his head while standing still. How much work does he do on the suitcase?
A 2 kg ball and a 4 kg ball are dropped from the same height (no air resistance). Just before hitting the ground, what is true about their speeds?
A ramp is made longer and gentler while still reaching the same platform height. What happens to the effort needed and the mechanical advantage?
Practice Problems
Try each one yourself before opening the solution.
Easy
A force of 25 N pushes a box 4 m across a floor in the direction of the force. How much work is done on the box?
Use W = F × s. Both force and displacement point the same way, so W = 25 N × 4 m = 100 J. The work done on the box is 100 joules.
What is the kinetic energy of a 0.5 kg ball moving at 4 m s⁻¹?
Use K = ½mv² = ½ × 0.5 kg × (4 m s⁻¹)² = ½ × 0.5 × 16 = 0.25 × 16 = 4 J. The ball has 4 joules of kinetic energy.
A 3 kg bag is lifted 2 m onto a shelf in 4 seconds. Find (a) the work done and (b) the power used. Take g = 10 m s⁻².
(a) Work against gravity: W = mgh = 3 kg × 10 m s⁻² × 2 m = 60 J.
(b) Power: P = W ÷ t = 60 J ÷ 4 s = 15 W. The work done is 60 J and the power used is 15 watts.
Medium
A student of mass 50 kg goes from the ground to the top of a building 72.5 m tall — first in a lift, then on another day by climbing the stairs. Take g = 10 m s⁻². (i) Find the gain in potential energy by the lift. (ii) Find the gain by climbing the stairs. (iii) What does this tell you about how potential energy depends on the path?
(i) Lift: gain in PE = mgh = 50 kg × 10 m s⁻² × 72.5 m = 36250 J.
(ii) Stairs: the start and end heights are exactly the same (ground to the same top, h = 72.5 m), so the gain in PE = mgh = 36250 J again.
(iii) The gain in gravitational potential energy is the same by both routes. So PE depends only on the change in height, not on the path taken to get there. The longer, winding stair route gives the same PE gain as the straight lift.
A 2 kg ball is thrown straight up with a velocity of 20 m s⁻¹. (i) What is the sign of the work done by gravity on the way up and on the way down? (ii) If the ball only reaches 19.4 m, how much work did air resistance do? Take g = 10 m s⁻².
(i) On the way up, gravity pulls down while the ball moves up — force and displacement are opposite, so gravity does negative work. On the way down, gravity pulls down while the ball moves down — same direction, so gravity does positive work.
(ii) Without air resistance, the ball’s kinetic energy at launch would all become potential energy at the top. Launch KE = ½mv² = ½ × 2 × 20² = ½ × 2 × 400 = 400 J. With no losses it would reach a height where mgh = 400 J, i.e. h = 400 ÷ (2 × 10) = 20 m. But it only reached 19.4 m. The PE actually gained = mgh = 2 × 10 × 19.4 = 388 J. The missing 400 − 388 = 12 J was removed by air resistance, so air resistance did −12 J of work.
A crane lifts a mass m to the 10th floor in a certain time t. The next time it lifts the same mass to the 20th floor (twice the height) in double the time, 2t. All floors have equal height. How do the energy and the power required compare?
Energy: lifting to twice the height needs twice the work, since W = mgh and h doubles. So the energy required doubles.
Power: P = W ÷ t. The work doubled (2W) but the time also doubled (2t). So new power = 2W ÷ 2t = W ÷ t, which is the same as before. The power required stays the same.
So: double the energy, but the same power.
Challenge
A 1.5 kg coconut falls from a tree 10 m tall onto wet sand. (i) Find its speed just before hitting the sand. (ii) If the sand pushes back with an average force of 3000 N and stops the coconut, find the depth of the depression it makes. Take g = 10 m s⁻².
(i) Use energy conservation as it falls: mgh = ½mv², so v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.1 m s⁻¹. (You can also leave it as v² = 200.)
(ii) When it hits the sand, all its energy goes into making the depression. The total energy delivered to the sand = the coconut’s potential energy from the full fall plus the depth d it sinks. The simplest route: all the energy from falling height (10 + d) is used up by the sand force over distance d.
Energy in = mg(h + d) = 1.5 × 10 × (10 + d) = 15 × (10 + d) = 150 + 15d.
Energy used by sand = force × depth = 3000 × d = 3000d.
Set them equal: 150 + 15d = 3000d, so 150 = 2985d, giving d ≈ 0.05 m ≈ 5 cm.
(If you ignore the small extra fall d during the depression, mgh = F × d gives 150 = 3000d, d = 0.05 m = 5 cm — the same answer to two figures. The depression is about 5 cm deep.)
On the Moon, gravity is about one-sixth of Earth's. An astronaut throws a ball straight up and it reaches 8 m on Earth. With the same throwing speed, how high will it go on the Moon?
At the throw, the ball has kinetic energy ½mv². At the top it has all become potential energy, mgh. So ½mv² = mgh, which gives h = v² ÷ (2g). The throwing speed v is the same on both worlds.
So h is inversely proportional to g: smaller g means a bigger height. On the Moon g is one-sixth of Earth’s, so the height becomes 6 times larger.
Moon height = 6 × 8 m = 48 m. The ball rises 48 m on the Moon.
Summary
You can now explain:
- Work in the scientific sense: W = F × s, the force times the distance moved in the direction of the force — and why holding a heavy bag still does zero work (no displacement).
- Why work can be positive (force and motion the same way) or negative (opposite), and that the unit of both work and energy is the joule.
- That energy is the capacity to do work, that doing work on an object changes its energy (the work-energy idea), and that energy comes in many forms that can switch from one to another.
- Kinetic energy K = ½mv² — and you can derive it from W = F × s and the kinematic equations — plus why doubling the speed gives four times the energy.
- Gravitational potential energy U = mgh, stored by lifting an object, and that potential energy can also be stored by deforming objects or separating them against a force.
- The conservation of mechanical energy: as something falls or swings, kinetic and potential energy trade off, but the total stays constant — and why a slide’s bottom speed v = √(2gh) does not depend on mass.
- Power P = W ÷ t, measured in watts, as the rate of doing work.
- How simple machines (pulley, inclined plane, lever) and their mechanical advantage (load ÷ effort) make tasks easier by trading force for distance — without ever reducing the total work or creating energy.
What’s Next
You have seen that energy can be stored in atoms — chemical energy in fuel and food, and nuclear energy locked deep inside the nucleus. But what is an atom made of, and how is all that energy held inside something so tiny? In Chapter 8 — Journey Inside the Atom, you will travel inside the atom to meet the electrons, protons and neutrons, and discover the surprising structure that holds them together.
Frequently Asked Questions
Why is holding a heavy bag still not 'work' in physics?
In science, work is force multiplied by the displacement in the direction of that force. If you hold a bag still, the bag does not move, so the displacement is zero. That makes the work done on the bag zero, no matter how heavy it is. You still feel tired because your muscles keep tightening and loosening and use up energy inside your body, but that is not work done on the bag.
How do you derive the formula ½mv² for kinetic energy?
Start with a constant force F pushing an object of mass m from rest through a distance s, giving it speed v. From kinematics, v² = u² + 2as, and with u = 0 this gives s = v² ÷ (2a). The work done is W = F × s = ma × s. Substituting s gives W = ma × v² ÷ (2a) = ½mv². By the work-energy idea, this work equals the energy the object gained, so kinetic energy = ½mv².
If a machine multiplies force, why doesn't it multiply energy too?
A simple machine cannot create energy. When a lever or ramp lets you push with a smaller force, you always have to move that force through a longer distance. Work is force times distance, so the two changes cancel out and the total work stays the same. The machine only trades a big force over a short distance for a small force over a long distance.
Why does a child's speed at the bottom of a slide not depend on their mass?
At the top, the child's potential energy is mgh. At the bottom this all turns into kinetic energy, ½mv². Setting mgh = ½mv², the mass m cancels from both sides, leaving v = √(2gh). So the speed depends only on the height of the slide, not on how heavy the child is. A heavy child and a light child reach the bottom of the same slide at the same speed (ignoring friction).
What is the difference between work and power?
Work is the total energy transferred when a force moves an object — measured in joules. Power is how fast that work is done — it is work divided by time, measured in watts. Walking up the stairs and running up the same stairs do the same work, but running needs more power because the same work is done in less time.
Why are mountain roads built winding around in gentle slopes instead of going straight up?
A winding road acts like a long, gentle inclined plane. A gentle slope needs a much smaller force to climb than a steep one, so vehicles can go up using less force from the engine. The trade-off is that the road is longer, so the vehicle travels a greater distance — but the total work against gravity stays the same. This makes the climb possible without an enormously powerful engine.