Class 9 Science Chapter 7: Work, Energy, and Simple Machines

The force applied multiplied by the displacement in the direction of that force, written $W = F \times s$. It holds for a displacement in any direction.

The SI unit of work, symbol J. One joule is done when a force of one newton shifts an object one metre along that force, which in base units is $1\ \text{kg m}^2\ \text{s}^{-2}$.

The work done equals the area enclosed between the plotted force and the displacement axis, and this holds even where the force is not constant.

Work is zero if the force is zero, if there is no displacement at all, or if the force acts at right angles to the displacement.

Displacement along the force gives positive work; displacement against it gives negative work. Work carries no direction, only a sign.

It is the force applied multiplied by the displacement in the direction of that force, written $W = F \times s$. The definition holds whether the displacement is vertical, horizontal or in any other direction.

Lifting three bags together to the same height needs a force three times as large over the same distance and gives three times the work. So a larger force over the same distance does proportionally more work. Lifting one bag three metres instead of one uses the same force over three times the distance and again gives three times the work, so the same force over a longer distance does proportionally more work too.

One joule of work is done when a constant force of one newton is applied to an object. It is displaced one metre in the direction of that force, so $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$. Since $1\ \text{N} = 1\ \text{kg m s}^{-2}$, it follows that $1\ \text{J} = 1\ \text{kg m}^2\ \text{s}^{-2}$.

The work equals the area enclosed under the plotted force between the initial and final positions. A steady 10 N over 1 m encloses a rectangle of 10 J. The method is useful because it still gives the work done when the force is not constant, which the simple product cannot handle.

The work is zero if the force itself is zero. It is zero if there is no displacement, as when a rigid wall is pushed and does not move. And it is zero if the force acts at right angles to the displacement, as when a box is carried horizontally while being held up.

Holding any push at all keeps the muscles tightening and slackening over and over, and that draws on the body's own store of energy. So you feel tired even though, in the scientific sense, no work has been done on the object, because there is no displacement.

Where something shifts the way the force points, the work that force does is positive, as with a wheelchair shoved forward. Where it shifts against the force, the work is negative, as when a keeper presses back on a ball still coming at her.

Her force opposes the ball's motion, so the displacement is taken as negative. The work is $200\ \text{N} \times (-0.15\ \text{m}) = -30$ J. Since the two also push on each other equally and oppositely, the ball does positive work on the goalkeeper at the same time.

Start with a 5 kg bag on the floor. To lift it slowly through one metre you must apply an upward force matching its weight, and the bag then moves one metre along that force. In everyday language you have done some work.

Now lift three such bags one after another to the same height. That is three times the task, so three times the work, and a machine doing it would need three times the fuel.

Lift all three together instead. You now apply a force three times as large over the same distance, and since the task is the same as before, the work is again three times that for one bag. So applying a larger force over the same distance does proportionally more work.

Finally lift a single bag to three metres. The same force now acts over three times the distance, and the work is once more three times that for one metre. So applying the same force over a larger distance does proportionally more work.

Both the force and the distance therefore count, each in proportion, which fixes the definition: the work done on an object by a constant force is the force applied multiplied by the displacement in the direction of that force, written $W = F \times s$. Its unit is the joule, with $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$.

All three follow from the definition, since work is the force multiplied by the displacement along that force.

The first is immediate. If the force acting on the object is zero, then no work is done on it, because there is nothing to multiply the displacement by.

The second is equally direct. If there is no displacement of the object, the work done on it is zero regardless of the force being applied. Applying a force to a rigid wall is the standard case: there is no displacement in the wall, so no work is done on the wall.

That case seems odd, because the person pushing feels tired. The explanation is that to apply a force the muscles in the body repeatedly expand and contract, using up the internal energy of the body. So tiredness is real even where, in the scientific sense, no work has been done on the object.

The third case is a force acting at right angles to the displacement. Here the work done by that force is zero because there is no displacement along the line of the force at all. Carry a box as you walk and your hands push up, holding its weight, while the box itself goes sideways — so that push does the box no work whatever.

Work done when the force and displacement meet at some other angle is left for higher grades.

Force and displacement both have magnitude and direction, but work does not. Work is described by a number carrying either a positive or a negative sign, and which sign it takes depends on the relative directions of the force and the displacement.

Where body and force go the same way, that force's work on the body counts as positive. Pushing a wheelchair is the standard case: the force and the displacement run the same way, so the work done on the wheelchair is positive.

Where the displacement is in the direction opposite to the force, the work done is negative. A goalkeeper stopping a ball applies a force against the ball's motion, and hence against its displacement, so she does negative work on the ball.

Taking the numbers, her hand moved back 15 cm while she applied 200 N. The displacement is taken as negative because the ball moves opposite to the applied force, so the work she does on the ball is $200\ \text{N} \times (-0.15\ \text{m}) = -30$ J.

One further point completes the picture. Each puts a force on the other, matched in size and opposed in sense, so each works on the other. What the ball does to the goalkeeper is positive work and what she does to the ball is negative — which is precisely why both parties have to be named every time.

Plot the force acting on an object up the vertical axis against the displacement in the direction of that force along the horizontal axis. The work done by the force is then the area enclosed under the plot between the initial and the final positions.

For a constant force the plot is a level line and the enclosed region is a rectangle. A steady force of 10 N over a displacement of 1 m gives an area of $10\ \text{N} \times 1\ \text{m} = 10$ J, which is exactly what the equation $W = F \times s$ would give.

So far the graph has added nothing. Its value appears once the force stops being constant.

The equation assumes a single fixed value of $F$ throughout, so it simply cannot be applied where the force changes as the object moves. The area, by contrast, can be read off whatever shape the plot takes, and it still gives the work done between the initial and final positions.

That is why the method is worth learning even in a section about constant forces: it is the general statement, of which the simple product is only the special case where the plot happens to be a level line.

The capacity to do work. An object gains it when positive work is done on it, and can spend it by applying a force to something else and setting it moving.

Work done on an object equals the change in its energy. It holds for a system of objects and even where the applied forces are not constant.

Mechanical, electrical, thermal, light, sound, nuclear and chemical. Energy can be converted from any one form into another.

The energy an object possesses on account of its motion or its position, which is the form most directly tied to forces and motion.

The energy an object possesses on account of its motion, given by $K = \frac{1}{2}mv^2$. It is measured in joules and has no direction.

Energy is the capacity to do work. A thrown ball knocks the wicket over and a raised flowerpot can damage what lies below, so each has acquired that capacity. The ball gained it from the work the fielder did in throwing it, and the pot from the work done in raising it.

The work done on an object equals the change in its energy. It holds not only for a single object but for a system of objects, and it holds even where the forces applied are not constant, which lets it settle problems that would otherwise be awkward.

Mechanical energy is due to the motion or position of objects. Thermal energy is what makes things warm or hot. Sound energy is the energy of vibrations of air or other molecules. Chemical energy is stored in fuels and food as bonds between atoms.

A bulb turns electrical energy into light, while an electric heater turns the same input into warmth in the water. What we eat carries chemical energy that our muscles turn into motion. And a bell that is struck turns motion into sound.

Take a constant force on a mass $m$ going from $u$ to $v$ over a displacement $s$. The acceleration is then constant, so $v^2 = u^2 + 2as$ gives $s = \frac{v^2 - u^2}{2a}$. Since $W = F \times s$ and $F = ma$, substituting gives $W = \frac{1}{2}m(v^2 - u^2)$, which by the theorem is the change in energy. Setting $u = 0$ leaves $K = \frac{1}{2}mv^2$.

When positive work is done the velocity increases and so does the kinetic energy. When negative work is done the velocity decreases and the kinetic energy decreases with it. When no work is done and the velocity is unchanged, the kinetic energy stays constant.

The initial kinetic energy is $\frac{1}{2}mv^2$. At double the velocity it becomes $\frac{1}{2}m(2v)^2 = \frac{1}{2} \times 4mv^2 = 4 \times \frac{1}{2}mv^2$. So it is four times the original value, because the velocity enters the formula squared.

Its final kinetic energy is zero, so the change is $-\frac{1}{2} \times 15000 \times v^2$. The wire's force opposes the displacement, so its work is $367500 \times (-100)$. Equating the two gives $v^2 = \frac{2 \times 36750000}{15000} = 4900$, so $v = 70\ \text{m s}^{-1}$, which is 252 km per hour towards the carrier.

Push something and shift it and work has been done to it, and the thing is left with something to show for it. A thrown cricket ball knocks over a wicket; a raised flowerpot can damage what lies below it. Each has acquired a capacity to do some work, and an object having that capacity is said to possess energy.

Where the energy came from is clear enough. The ball gained it from the work the fielder did in throwing it, and the pot from the work done in raising it. When positive work is done upon an object it gains energy, which it can then spend by applying a force to another object and setting it moving, so passing the energy on.

That relation is the work-energy theorem: the work done on an object equals the change in its energy. It holds for a system of objects as well as a single one, and even where the applied forces are not constant.

A carrom shot shows it working twice over. The moving striker applies a force in the direction of the white coin's displacement, so it does positive work on the coin and the coin's energy increases. The third law returns an opposite push from coin to striker, which is negative work on the striker, and its energy falls.

The same exchange repeats at the next collision, the white coin gaining the black coin energy and losing some of its own.

Kinetic energy is the energy an object possesses due to its motion, and an object that does not move is defined to have none. For a formula, let a body of mass $m$ set out already moving at $u$ rather than from standstill, and finish at $v$ after travelling $s$.

Assume the force $F$ acting on it is constant. Then the acceleration $a$ is constant too, so the kinematic equations apply, and in particular $v^2 = u^2 + 2as$, which rearranges to $s = \frac{v^2 - u^2}{2a}$.

From the definition of work, $W = F \times s$, and using Newton's second law $F = ma$ this becomes $W = ma \times s$. Substituting for $s$ gives $W = ma \times \frac{v^2 - u^2}{2a} = \frac{1}{2}m(v^2 - u^2)$, the acceleration cancelling out.

By the work-energy theorem, the work done by the force translates into the change in the object's energy, so that expression is the change.

Now set the initial velocity to zero. The change in energy is then simply the final kinetic energy, giving $K = \frac{1}{2}mv^2$ for a mass $m$ moving at velocity $v$. Its unit is the joule, and it has no direction at all.

Energy is the capacity to do work, and it exists in many forms. Mechanical energy is due to the motion or position of objects. Electrical energy is related to the position or motion of charges. Thermal energy is what makes things warm or hot, light energy is what allows us to see, and sound energy is the energy of vibrations of air or other molecules. Nuclear energy is stored in the nuclei of atoms, and chemical energy is stored in fuels and food as bonds between atoms.

Any one form can be changed into another. Electrical energy becomes light in a bulb and thermal energy of the water in an electric heater, the chemical energy of food powers our muscles as mechanical energy, and a ringing bell turns mechanical energy into sound.

As for moving energy about, doing mechanical work is one way but not the only one. Energy travels as heat: when two bodies at unequal temperatures come into contact, it flows from the hotter to the colder.

It can also move without any contact at all. The Sun's energy reaches the Earth through radiation.

Besides these, energy is transferred in electric circuits, through sound waves, and in the nuclear reactions that power the Sun itself.

The aircraft arrives with some velocity $v$ and is brought to rest, so the whole of its kinetic energy has to be removed by the wire. That makes it a work-energy problem rather than a kinematics one.

Its initial kinetic energy is $\frac{1}{2} \times 15000\ \text{kg} \times v^2$, and its final kinetic energy is zero, so the change in kinetic energy is $0 - \frac{1}{2} \times 15000 \times v^2$.

By the work-energy theorem that change equals the work done by the wire. The aircraft's displacement and the force the wire applies are in opposite directions, so the work done by the wire is negative, namely $367500\ \text{N} \times (-100\ \text{m})$.

Setting the two equal gives $-\frac{1}{2} \times 15000 \times v^2 = -367500 \times 100$, so that $v^2 = \frac{2 \times 36750000}{15000} = 4900\ \text{m}^2\ \text{s}^{-2}$.

Taking the square root, $v = 70\ \text{m s}^{-1}$, which is 252 km per hour, directed towards the aircraft carrier. Notice what the theorem achieved: a landing speed obtained from nothing more than a force and a distance.

Energy stored either by an object's deformation, or in a system of objects on account of their relative positions.

An external force overcomes the internal forces of a spring or band to change its shape. When released, those internal forces undo the deformation and do work.

Separating two magnets, two charges or a ball from the Earth stores energy. Released, the objects rush together and gain kinetic energy.

The stored energy of the Earth-object system, usually spoken of as the object's alone since the Earth scarcely moves. It is given by $U = mgh$.

A matter of definition. Here the potential energy of an object lying on the ground is taken as zero, and $mgh$ measures the gain from raising it.

When the bow is released it applies a force on the arrow in contact with it, setting the arrow in motion and giving it kinetic energy, while the bow itself returns to its original shape. That energy came from the bow, which had a capacity to do work on account of its shape. The work done by the archer in bending it was stored there.

Stretching or compressing a spring is work done on it to change its shape, and while it is held deformed it stores energy on account of its configuration. An external force is needed to overcome the internal forces within the spring in order to deform it. Once that external force is removed, the internal forces undo the deformation and carry out work, handing the stored energy over as kinetic energy to whatever the spring touches.

By changing the arrangement of the objects. Separating the unlike poles of two magnets takes a force, and releasing them sends them rushing together with enough kinetic energy to crush a pea between two strong ones. A pair of separated electric charges stores energy the same way, and so does a ball lifted away from the Earth.

Potential energy is whatever a body holds through having been deformed, or a group of bodies holds through how far apart they have been set. As a general rule, any group of bodies that pull or push on one another by gravity, electricity or magnetism can hold energy according to how they are arranged.

A heavy ball is dropped onto a bed of loose sand from about 1 m, and then from 2 m at a spot where the hollows will not overlap, the depths being compared. The depression is deepest for the drop from the greatest height. Raising the ball higher takes more work, so it holds more energy, and releasing it from there digs deeper. So the greater the height above the Earth's surface, the greater the potential energy.

Take an object of mass $m$ lying on the ground and define its potential energy there to be zero. Raising it gradually to a height $h$ needs a force equal to $mg$, the force between the Earth and the object, so the work done is $W = mg \times h = mgh$. By the work-energy theorem that work appears as a change in the object's potential energy, giving $U = mgh$.

First convert the mass: 200 g is 0.2 kg. Then $U = mgh = 0.2\ \text{kg} \times 10\ \text{m s}^{-2} \times 10\ \text{m} = 20$ J.

Moving horizontally at constant velocity does not change it, because $U = mgh$ depends on the height alone and the height has not altered. Being gradually raised does change it, since $h$ increases and the potential energy rises with it.

The first is by deforming an object. Release a stretched slingshot band or a drawn bow and it applies a force on whatever it touches, setting that object moving and giving it kinetic energy, while the band or bow returns to its original shape. The energy came from the deformed object, which had a capacity to do work on account of its shape, and what was stored there was the work done in deforming it. A spring behaves the same way, holding energy by its configuration while deformed and transferring it as kinetic energy on release.

Deforming the object means applying an external force to overcome the internal forces within it. When the external force is withdrawn, those internal forces undo the deformation and can carry out work, which is how internal forces let energy be stored inside a deformed object.

The second way is by changing the arrangement of objects in a system, with no deformation at all. Separating the unlike poles of two magnets takes a force, and releasing them sends them rushing together with kinetic energy enough to crush a pea. A pair of separated charges stores energy the same way, and so does a ball lifted away from the Earth, since the two attract and rush together when released.

Generally, wherever a system interacts through gravitational, electric or magnetic forces, it can store energy by the relative positions of its objects.

The apparatus is a heavy ball and a big container holding loose sand.

The ball is first raised about 1 m above the sand bed and dropped, and a depression appears where it lands. It is then raised to 2 m and released at a slightly different position, chosen so that the depressions do not overlap and their depths can be compared fairly. The step is repeated once more.

Set the hollows side by side and the pattern is plain: the higher the drop the deeper the mark, the lower the drop the shallower.

The explanation runs through work. Carrying a ball further up from the ground costs more work than carrying it a shorter way. The ball therefore possesses more energy at the greater height, and when released from there, that energy is what creates the deeper depression in the sand.

The conclusion is immediate: every metre further from the ground leaves the ball with that much more potential energy. The activity does not measure the energy, but it does show it rising with height, which is exactly what the expression $U = mgh$ later states in symbols.

Consider an object of mass $m$ lying on the ground. We are free to define the potential energy to be zero in this configuration, and doing so fixes the reference from which heights are measured.

To raise the object gradually from the Earth's surface to a height $h$, a force equal to $mg$ must be applied, that being the force between the Earth and the object. Raising it gradually matters, because a larger force would also give the object kinetic energy.

The work done by that applied force is therefore $W = \text{force} \times \text{displacement} = mg \times h = mgh$.

The work-energy theorem turns that work into energy the body did not have before; with nothing moving at either end of the lift, the whole of it has gone into potential energy. So the potential energy of the object at height $h$ is $U = mgh$, measured in joules like work and kinetic energy.

Applying it: a fielder throws a cricket ball of mass 200 g about 10 m up, with $g$ taken as $10\ \text{m s}^{-2}$. Converting the mass to 0.2 kg, the potential energy at the top is $U = 0.2 \times 10 \times 10 = 20$ J.

Strictly, the energy is stored by a system of two bodies. A ball lying on the Earth's surface and the Earth itself attract each other, so lifting the ball means doing work against the gravitational force between them, and when the lifting force is removed the two rush towards each other and gain kinetic energy. The store belongs to the pair.

But the Earth is very much more massive than the ball. Just as with the gun and the bullet in Chapter 6, the same force produces a far smaller acceleration in the more massive body, so the Earth hardly moves towards the ball at all.

For that reason the stored energy of the Earth-ball system is usually referred to simply as the gravitational potential energy of the ball, and from that point on in the chapter potential energy generally means the gravitational kind.

One limit is worth carrying forward. Work done on a system against its internal forces can result in a gain of potential energy where those forces are gravitational, electric or magnetic.

That is not true of every internal force. Work done against friction does not lead to any storage of energy at all, and learning to tell which forces behave which way is left for higher grades.

The sum of a body's kinetic and potential energies. For a body moving under gravity alone it stays constant throughout the motion.

As a body moves under the gravitational force, its mechanical energy remains the same, provided no other external force acts on it.

Friction at the support and air resistance carry energy away, so the bob falls slightly short of its starting height on each swing and eventually halts.

The rate at which work is done. The average power is the work divided by the time taken, $P = \frac{W}{t}$, measured in watts.

One joule of work done per second, so $1\ \text{W} = 1\ \text{J s}^{-1}$. One horsepower is 746 W.

It is the sum of the kinetic and potential energies. A mass released from height $h$ starts with $mgh$ and no motion, so its mechanical energy is $mgh$. After falling for a time $t$ its potential energy is $mgh - \frac{1}{2}mg^2t^2$ and its kinetic energy is $\frac{1}{2}mg^2t^2$, which add back to $mgh$. The kinetic energy has gained exactly what the potential energy lost, so the sum is unchanged.

Let gravity alone move a body, with nothing else acting on it from outside, and the total it carries never alters. That proviso matters, since friction and air resistance are external forces and do carry energy away.

A simple pendulum is set up with paper on the wall behind and a horizontal line drawn above the resting bob. The bob is taken aside to a point level with that line and released. Over the first couple of oscillations it almost reaches the level of the line on the far side, which demonstrates that its mechanical energy remains constant.

At the release point it has potential energy $mgh$, measured from the lowest point, and no kinetic energy. At the lowest point its potential energy is zero and all the energy is kinetic. At the far extreme, where it stops, the kinetic energy is zero again and the potential energy has been regained.

Because energy is lost to friction at the support and to air resistance. Both are external forces, and conservation of mechanical energy was only ever claimed for the case where no such force acts.

The potential energy at the top, $mgh$, is converted entirely into kinetic energy at the bottom, $\frac{1}{2}mv^2$, if friction is neglected. Equating them and cancelling the mass gives $v = \sqrt{2gh}$. Since only the height appears, the answer depends neither on the shape of the slide nor on the mass of the child.

Power is the rate at which work is done, the average power being $P = \frac{W}{t}$, the work done divided by the time taken. Its SI unit is the watt, where one watt is one joule of work done per second, so $1\ \text{W} = 1\ \text{J s}^{-1}$.

Converting, 72 km h⁻¹ is 20 m s⁻¹. The work done by the engine is the gain in kinetic energy, $\frac{1}{2} \times 1000 \times 20^2 - 0 = 200000$ J. The power is therefore $\frac{200000}{10} = 20000$ W.

Mechanical energy is the sum of an object's kinetic and potential energies, and the question is what happens to that sum during a free fall.

Take a mass $m$ lifted to point A at height $h$ and dropped, so its initial velocity is zero. Its potential energy there is $mgh$ and its kinetic energy is nothing, so its mechanical energy is $mgh$.

After falling under gravity for a time $t$ it reaches point B. From the kinematic equations, with the acceleration taken as $g$, its velocity is $v = gt$ and its height is $h' = h - \frac{1}{2}gt^2$.

At B, therefore, the potential energy is $mgh' = mgh - \frac{1}{2}mg^2t^2$, while the kinetic energy is $\frac{1}{2}mv^2 = \frac{1}{2}mg^2t^2$. Adding them, the term $\frac{1}{2}mg^2t^2$ cancels and the mechanical energy is $mgh$ once more.

Comparing the two points shows what happened. The kinetic energy gained exactly $\frac{1}{2}mg^2t^2$, which is exactly what the potential energy lost, so the lost potential energy was converted into kinetic energy while the total stayed fixed.

So where gravity alone moves a body and nothing else acts on it from outside, the total it carries never alters.

A simple pendulum is set up, a white sheet of paper is pasted on the wall behind it, and a horizontal line is drawn above the position of the bob when it hangs at rest. The bob is then taken to one side, to a point P level with that line, and released, and the extreme points of the first couple of oscillations are watched.

At P the bob has potential energy $mgh$, where $h$ is measured from the lowest point of its path, and no kinetic energy at all, since it starts from rest.

At Q, the lowest point, the potential energy is zero but the bob is moving fastest, so all of the energy is kinetic.

At R, on the far side where the bob stops, the kinetic energy is zero again and the potential energy has been regained. The bob reaches almost the same height it started from, and that is what demonstrates that its mechanical energy remains constant.

The word almost matters. In real life a pendulum slows and finally comes to rest, because energy is lost to friction at the support and to air resistance. Both are external forces, so the condition on which conservation rests is not quite met.

Escape ramps are inclined planes filled with sand or gravel, built to stop trucks whose brakes fail on a highway. Here a 10000 kg truck is travelling at 72 km h⁻¹ when its brakes fail, and it is steered onto a ramp inclined at 30 degrees where the sand resists with 50000 N. The question is the minimum ramp length.

First convert: 72 km h⁻¹ is 20 m s⁻¹. The truck's initial kinetic energy is $\frac{1}{2} \times 10000 \times 20^2 = 2000000$ J, and its initial potential energy is zero, so its total initial energy is 2000000 J.

Let it travel a distance $d$ along the ramp. On a 30 degree incline it rises 1 m for every 2 m travelled, so the height gained is $\frac{d}{2}$. Its final kinetic energy is zero, and its final potential energy is $mg \times \frac{d}{2} = 50000d$.

The change in total energy is therefore $50000d - 2000000$, while the work done by the sand is negative, being $-50000d$, since the force opposes the motion.

By the work-energy theorem those are equal, so $-50000d = 50000d - 2000000$, giving $2000000 = 100000d$ and $d = 20$ m. Notice that the rise and the sand each account for half of the stopping.

Carrying a bag up a flight of stairs by running in one minute feels very different from walking up slowly in five, even though the same work is done in both cases. Work alone cannot tell the two apart, and power is the quantity that does.

Power is the rate at which work is done. The average power is the work done divided by the time taken, $P = \frac{W}{t}$. Two consequences follow at once: doing more work in the same time requires more power, and doing the same work in a shorter time also requires more power.

The SI unit is the watt, one watt being one joule of work done per second, so $1\ \text{W} = 1\ \text{J s}^{-1}$. Another unit, horsepower, is used for car engines and water pumps, one horsepower being 746 W, a name dating from when early engines were compared with the horses that drew carriages.

For a weightlifter raising 75 kg through 2 m in 5 s, the work is $mgh = 75 \times 10 \times 2 = 1500$ J, so the power is $\frac{1500}{5} = 300$ W.

For a 1000 kg car reaching 20 m s⁻¹ from rest in 10 s, the work is the gain in kinetic energy, $\frac{1}{2} \times 1000 \times 400 = 200000$ J, so the power is $\frac{200000}{10} = 20000$ W.

A device that makes a task easier by changing the magnitude or direction of the force required. It cannot reduce the total work the task demands.

The ratio of the load to the effort. A value greater than 1 means the machine delivers more force to the load than is applied to it.

A grooved wheel guiding a rope. It changes only the direction of the effort, not its size, so its mechanical advantage is 1.

A simple machine for moving a heavy load to a higher or lower level. Its mechanical advantage is $\frac{L}{h}$, the length divided by the height.

A rigid bar free to turn about a fixed point called the fulcrum. It balances when effort times effort arm equals load times load arm.

A simple machine is a device that makes a task easier by altering either the magnitude or the direction of the force that must be applied. What it cannot do is reduce the total work the task requires, since work is force multiplied by displacement and a smaller force simply acts over a longer distance.

The effort is the force we apply to a machine, the load is the force that has to be overcome. The mechanical advantage is the ratio of the load to the effort.

It does not reduce the magnitude of the force required, only its direction, so its mechanical advantage is 1 since effort and load are equal. It is still worth using because pulling downward is easier for us than applying an upward force directly, so it provides convenience.

A cart with a spring balance attached is first lifted vertically to a raised surface, the reading giving its weight. It is then pulled up a smooth plank set against that surface, and the reading is smaller. Reducing the angle so the plank is less steep reduces the reading further, but the force must then be applied over a larger distance to reach the same height.

With the object of mass $m$ moved at constant speed up a plane of length $L$ to a height $h$, the work done is $F' \times L$ and the potential energy gained is $mgh$. Ignoring friction, the work-energy theorem gives $F' \times L = mgh$, so $\frac{mg}{F'} = \frac{L}{h}$. Since $mg$ is the load and $F'$ the effort, the mechanical advantage is $\frac{L}{h}$.

By the right-angled triangle property, a height of 30 cm and a width of 40 cm give a sloping length of 50 cm. The mechanical advantage is therefore $\frac{L}{h} = \frac{50}{30} = 1.67$.

A lever is a rigid bar that rotates about a fixed point. Its parts are the fulcrum, the fixed point it turns about; the load, the force to be overcome; and the effort, the force applied. The distances from the fulcrum are the load arm and the effort arm, and the lever balances when effort multiplied by effort arm equals load multiplied by load arm.

In Class I the fulcrum lies between the effort and the load, as in scissors or a seesaw. In Class II the load lies in between, as in a wheelbarrow or bottle opener. In Class III the effort lies in between, as in tweezers or a broom.

Jobs done against gravity or against any other force share one awkward feature: the work they demand is fixed. What can be altered is how large the force must be, or which way it has to point, and the devices that alter it are called simple machines.

Three terms make the discussion precise. The force we apply is the effort, the force to be overcome is the load, and the mechanical advantage is the ratio of the load to the effort.

A pulley is a grooved wheel that keeps a rope running true. Fixed in place, a pulley leaves the force exactly as large as it was, and alters only which way you must pull. Since the effort and the load are equal in magnitude, its mechanical advantage is 1.

That is not useless. Hauling down suits our bodies far better than hoisting a weight straight up with our arms, so what the fixed pulley offers is convenience and not strength — which is why flags and loads go up on one.

Movable pulleys, or systems of several, can have a mechanical advantage greater than 1 and lift much heavier objects with much smaller effort. The arrangement hangs the load from the travelling pulley; of the rope's two ends, one is made fast and the other is where you haul. Elevators and cranes make use of them.

The apparatus is a smooth plank about 1.5 m long, a cart with a spring balance attached, and a raised surface such as a stool or a pile of books about half a metre above the floor.

First the cart is lifted vertically, slowly and steadily, to the top, and the spring balance reading is noted. That reading is the force required to lift it vertically, which equals the weight of the cart. Next the plank is placed against the top and the cart pulled along it, slowly and steadily, and the reading proves smaller. Reducing the angle so the plank is less steep reduces it further still, although the force must then be applied over a larger distance to reach the same height.

To find the mechanical advantage, let the object's mass be $m$, so the load is its weight $mg$, and let $F'$ be the effort needed to raise it to a height $h$ along a plane of length $L$.

Moving at constant speed, the total work done on the object is $F' \times L$, while the potential energy it gains is $mgh$. By the work-energy theorem, ignoring friction, these are equal, so $F' \times L = mgh$ and therefore $\frac{mg}{F'} = \frac{L}{h}$.

Since $mg$ is the load and $F'$ the effort, the mechanical advantage is $\frac{L}{h}$, and because $L$ exceeds $h$ it is always greater than 1.

A lever is a rigid bar that can rotate about a fixed point. Three things make it up: the fulcrum, that fixed point; the load, which is what must be overcome; and the effort, which is what you supply. Measure from the fulcrum out to the load and you have the load arm; measure from the fulcrum out to where you push and you have the effort arm.

A small force at one end can apply a large force at the other because the end taking the smaller force $F_1$ moves through a larger distance $d_1$, while the end delivering the larger force $F_2$ moves through a smaller distance $d_2$. The work done on one end is transferred to the other, so $F_1 \times d_1 = F_2 \times d_2$.

Activity 7.5 makes this measurable. A long scale is tied at its midpoint with string, which serves as the fulcrum, and hung so it can swing freely as a beam. Paper cups are fixed at both ends as pans and the beam levelled. One coin in each pan leaves it horizontal; adding a second coin to one pan tilts it, and that pan must be moved closer to the centre to restore balance, the distance being measured each time for 2, 4 and 8 coins.

Analysing the readings gives the principle: effort multiplied by effort arm equals load multiplied by load arm, so the mechanical advantage is the effort arm divided by the load arm.

Levers fall into three classes according to what lies between the other two parts. In Class I the fulcrum is in between, as with tongs, scissors, a crowbar, pliers, a balance scale or a seesaw. In Class II the load is in between, as with a lemon squeezer, a wheelbarrow or a bottle opener. In Class III the effort is in between, as with tongs, tweezers, a broom, a hammer or an oar.

The seesaw example applies the principle directly. With seats at 2 m and 1 m from the fulcrum on each side, suppose the 15 kg child sits at the far seat and the other a distance $L$ from the fulcrum. Then $15 \times 2 = 30 \times L$, giving $L = 1$ m, so the 30 kg child should take the nearer seat.

A general point closes the chapter. A lever usually asks a smaller push of you, but it asks that push over a longer travel, and the work you end up doing is neither more nor less than before. The same trade appears in every simple machine.

In every case the conservation of mechanical energy still holds: the work put in equals the useful work done on the load, once friction is ignored.

So no machine makes energy; what a machine does is let us spend ours to better purpose — which is why every perpetual contrivance of wheels, weights or magnets ever proposed has failed.

It is the force applied multiplied by the displacement in the direction of that force, written $W = F \times s$. The definition holds whether the displacement is vertical, horizontal or in any other direction.

Lifting three bags together to the same height needs a force three times as large over the same distance and gives three times the work. So a larger force over the same distance does proportionally more work. Lifting one bag three metres instead of one uses the same force over three times the distance and again gives three times the work, so the same force over a longer distance does proportionally more work too.

One joule of work is done when a constant force of one newton is applied to an object. It is displaced one metre in the direction of that force, so $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$. Since $1\ \text{N} = 1\ \text{kg m s}^{-2}$, it follows that $1\ \text{J} = 1\ \text{kg m}^2\ \text{s}^{-2}$.

Energy is the capacity to do work. A thrown ball knocks the wicket over and a raised flowerpot can damage what lies below, so each has acquired that capacity. The ball gained it from the work the fielder did in throwing it, and the pot from the work done in raising it.

The work done on an object equals the change in its energy. It holds not only for a single object but for a system of objects, and it holds even where the forces applied are not constant, which lets it settle problems that would otherwise be awkward.

Mechanical energy is due to the motion or position of objects. Thermal energy is what makes things warm or hot. Sound energy is the energy of vibrations of air or other molecules. Chemical energy is stored in fuels and food as bonds between atoms.

When the bow is released it applies a force on the arrow in contact with it, setting the arrow in motion and giving it kinetic energy, while the bow itself returns to its original shape. That energy came from the bow, which had a capacity to do work on account of its shape. The work done by the archer in bending it was stored there.

Stretching or compressing a spring is work done on it to change its shape, and while it is held deformed it stores energy on account of its configuration. An external force is needed to overcome the internal forces within the spring in order to deform it. Once that external force is removed, the internal forces undo the deformation and carry out work, handing the stored energy over as kinetic energy to whatever the spring touches.

By changing the arrangement of the objects. Separating the unlike poles of two magnets takes a force, and releasing them sends them rushing together with enough kinetic energy to crush a pea between two strong ones. A pair of separated electric charges stores energy the same way, and so does a ball lifted away from the Earth.

It is the sum of the kinetic and potential energies. A mass released from height $h$ starts with $mgh$ and no motion, so its mechanical energy is $mgh$. After falling for a time $t$ its potential energy is $mgh - \frac{1}{2}mg^2t^2$ and its kinetic energy is $\frac{1}{2}mg^2t^2$, which add back to $mgh$. The kinetic energy has gained exactly what the potential energy lost, so the sum is unchanged.