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

Exploration · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

100 min
31 practice questions
31 MCQs

What you will learn in Work, Energy, and Simple Machines

Calculate the work done by a constant force with W = F × s, and decide when work is positive, negative or zero
Find work as the area under a force–displacement graph
State the work–energy theorem and use it to link work done to a change in energy
Name the main forms of energy and trace energy changes from one form to another
Derive and use K = ½mv² and U = mgh, and apply conservation of mechanical energy to falls, slides and pendulums
Calculate average power with P = W/t and use the watt
Explain how pulleys, inclined planes and levers make tasks easier, and calculate their mechanical advantage

Original companion text aligned to NCERT Science Class 9, Chapter 7, pp. 116–139. Not affiliated with or endorsed by NCERT.

At a water park, three slides start from the same tower. One plunges almost straight down, one sweeps round in a smooth curve, and one twists in a long spiral. Riders shoot out of all three into the same pool. Which slide gives the fastest splash? Does a heavier rider come out faster than a lighter one?

You could attack these questions with forces and the equations of motion from Chapters 4 and 6. But on a curving slide the push on the rider changes direction every moment, and the working soon gets heavy. This chapter gives you a shortcut: the ideas of work, energy and power. Instead of following every detail of a motion, you keep an account of energy, rather like keeping track of the money in a wallet.

Every activity draws its energy from somewhere. The dal-chawal you ate fuels your legs when you cycle to school. Electricity spins the blades of a mixer. Burning LPG heats the pressure cooker. Later in the chapter you will meet simple machines: pulleys, ramps and levers that make heavy jobs feel lighter, without ever handing you free energy.

7.1 Work Done by a Constant Force

In everyday talk, work is anything that takes effort: homework, a long shift at a shop, even standing in a queue. Science uses the word far more narrowly. A force does work (कार्य) on an object only when the object moves along the line of that force. For now we keep things simple: the force stays constant, and the object moves in a straight line along it.

How should work be measured? Think of a railway porter who is paid by the job. Lifting a suitcase twice as heavy onto the same rack is clearly twice the job. So is carrying the same suitcase up two flights of stairs instead of one. Science measures work in the same fair way: it grows in proportion to the force, and in proportion to the distance moved along the force.

Definition

Work done by a constant force

The force multiplied by the displacement of the object in the direction of the force.

ExampleA steady $50 N$ push that moves a cart $4 m$ forward does $200 J$ of work on the cart.

Work done by a constant force

W = F \times s

(7.2)

where

$W$: work done (joules, J)
$F$: the constant force (newtons, N)
$s$: displacement along the force (metres, m)

The force and the displacement can point any way: along a floor, straight up from a shelf, down a ramp. What matters is that they lie along the same line. When you describe work, always name the force (or the agent) doing it and the object it acts on: the porter's lifting force does work on the suitcase.

The SI unit of work is the joule (J). One joule is the work done when a force of $1 N$ moves an object $1 m$ along the force, so $1 J = 1 N m .$ Since $1 N = 1 k g m s^{- 2},$ a joule is also $1 k g m^{2} s^{- 2} .$ To get a feel for its size: lifting a $100 g$ apple through $1 m$ takes about one joule.

A graph makes the idea easy to see. Plot force on the vertical axis against displacement on the horizontal axis. A constant force gives a flat line, and the work done is the area of the rectangle under it: height × width = force × displacement (Fig. 7.1). The same rule holds when the force changes as the object moves. The work done is still the area under the force–displacement graph, whatever its shape.

2/4 $400 N$ over the same $2 m$ : twice as tall, so twice the work, $800 J .$

Fig. 7.1The area under a force–displacement graph equals the work done.NCERT Fig. 7.4 · p. 118

A graph of force in newtons, from 0 to 400, against displacement in metres, from 0 to 4. In the first step a rectangle 200 N high and 2 m wide is shaded and labelled 400 J. In the second step the rectangle is 400 N high and 2 m wide, labelled 800 J, with the first rectangle shown as a dashed outline. In the third step it is 200 N high and 4 m wide, also 800 J. In the last step a triangle rises from 0 N to 400 N over 3 m, labelled 600 J.

7.1.1 When is work done equal to zero?

Look again at $W = F \times s .$ The product is zero if either factor is zero. There is one more case hidden in the words “along the force”. So a force does no work in three situations (Fig. 7.2).

No displacement ( $s = 0$ ). Push with all your strength on a loaded truck parked with its brakes on. The truck does not budge, so your push does no work on it.
Force at right angles to the motion. A waiter carries a tray across a hall at a steady height. His hand pushes up on the tray, but the tray moves sideways. That upward force does no work on the tray.
No force along the motion ( $F = 0$ ). A hockey puck gliding over smooth ice keeps moving, yet nothing pushes it forward. No work is done on it as it glides.

Why, then, do you feel worn out after pushing the truck? To keep up the push, your muscles tighten and relax many times every second. That uses energy from your food, which ends up warming your body. You have used energy, but you have done no work on the truck. Finding the work done by a force that acts at some other angle needs a little trigonometry, which comes in higher classes.

2/3The upward force on the tray is at right angles to its motion, so it does no work.

Fig. 7.2Three ways a force can do no work.NCERT Figs. 7.5, 7.6 · p. 118

Three scenes, one per step. First, a person leans on the back of a large parked truck; a force arrow points into the truck, and the label says s = 0 because the truck does not move. Second, a waiter holds a tray level; a separate sketch shows an upward force arrow F and a sideways displacement arrow s meeting at a right-angle mark labelled 90 degrees. Third, a hockey puck glides along a strip of ice with a displacement arrow, and the label says F = 0 along s.

7.1.2 Positive and negative work done

Force and displacement both have directions. Work combines them into a single number with no direction, but it does have a sign: positive, negative or zero.

If the object moves the same way as the force acts, the force does positive work. A shopper pushing a loaded supermarket trolley forward does positive work on it. If the object moves opposite to the force, the force does negative work. A railway porter who stops a rolling luggage trolley pushes backwards on it while it is still moving forwards, so he does negative work on the trolley (Fig. 7.3).

Forces come in pairs (Newton's third law). While the porter pushes back on the trolley, the trolley pushes forward on his hands, and his hands give way a little in that forward direction. So the trolley does positive work on the porter at the very moment the porter does negative work on the trolley.

2/3The porter's force points backwards while the trolley moves forwards: negative work on the trolley.

Fig. 7.3The sign of the work depends on whether force and displacement point the same way.NCERT Fig. 7.7 · p. 119

A luggage trolley loaded with two boxes sits on a platform with a dashed displacement arrow s pointing right. In the first step a shopper behind it pushes, and the force arrow F also points right: positive work. In the second step a porter stands in front of it; the force arrow F points left, against the displacement: negative work. In the third step a green arrow shows the trolley pushing the porter's hands forward, with labels: on trolley minus 60 J, on porter plus 60 J.

Example 7.1NCERT Example 7.1 · p. 119

Up and down the well

A girl draws a bucket of water up from a village well, pulling hand over hand on the rope. Later she lowers an empty bucket back down slowly, holding the rope back as it slides through her hands. In each case, is the work done by her pull on the bucket positive or negative? What about the work done by gravity?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.2NCERT Example 7.2 · p. 119

Stopping a luggage trolley

A porter brings a rolling luggage trolley to rest by pushing back on it with a steady force of $120 N .$ The trolley moves $0.5 m$ forward while it stops. How much work does the porter do on the trolley? How much does the trolley do on the porter?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Pause and ponder: questions with answers

Short answerEasyPause and Ponder Q1
A coolie stands still on a platform for two minutes, balancing a $20 k g$ trunk on his head while he waits for a train. How much work does he do on the trunk? Why does he still get tired?
Show solution
AnswerZero work on the trunk; his muscles still use energy to hold it.
The trunk does not move, so $s = 0$ and $W = F \times 0 = 0.$ His upward force does no work on it. His neck, back and leg muscles must keep tightening to hold the load. That uses energy from his food, which turns into heat in his body.
Quick answerEasyPause and Ponder Q2
You shove a book across a table. After it leaves your hand it slides and stops. While it slides, what kind of work does friction do on the book?
positive worknegative workzero work
Show answer
Answer (B) negative work
Friction acts backwards while the book moves forwards. Force and displacement are opposite, so friction does negative work, which is why the book loses speed.

Quick check

1 of 4

A vendor pushes his handcart along a level road with a steady force of

150 N

for

40 m

in the direction of the push. How much work does his push do on the cart?

Show answer

Answer 6000 J

W = 150 N \times 40 m = 6000 J .

7.2 The Work-Energy Theorem

A swinging hammer drives a nail into a plank. Water stored in a rooftop tank gushes out of a tap far below. A wound-up toy car whirrs across the floor. Each of these can push something else and move it: each can do work. Anything that can do work has energy (ऊर्जा). Energy is the capacity to do work.

Where did that capacity come from? In every case something first did work on the object. Your arm did work on the hammer as you swung it. A pump did work raising the water to the tank. Your fingers did work winding the spring. Positive work done on an object increases its energy. The object can later do work on something else and pass the energy on. When the hammer hits the nail it drives the nail into the wood and slows down: its energy goes into the nail and the wood (Fig. 7.4).

2/3The hammer pushes the nail into the wood (positive work on the nail) while the nail pushes back on the hammer (negative work on it).

Fig. 7.4Work passes energy from your arm to the hammer, and from the hammer to the nail.NCERT Fig. 7.8 · p. 120

A nail stands upright in a wooden plank, and a hammer is held above it. In the first step the hammer is raised, with a dashed arc showing its swing, and the labels read: arm does work, hammer gains kinetic energy. In the second step the hammer head has come down on the nail and driven it deeper into the wood; labels read plus work on nail and minus work on hammer. In the third step the hammer rests on the nail, and the label says the energy is now in the nail, the wood and heat.

TheoremWork-energy theorem

The work done on an object equals the change in its energy.

W = E_{final} - E_{initial}

(7.3)

where

$W$: work done on the object (J)
$E_{initial}, E_{final}$: its energy before and after (J)

This is a very general result. It works for forces that vary as the object moves, and for whole systems of objects, such as a train of coupled wagons. And because work turns into energy joule for joule, energy is measured in the same unit as work: the joule.

Example 7.3NCERT Example 7.3 · p. 120

A chain of marbles

In a game of marbles (kanche), Ravi flicks his shooter marble with a finger. It rolls into a second marble, which rolls on and strikes a third. Say who does work on whom at each stage, and how each marble's energy changes.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 1 question

Pause and ponder: questions with answers

Short answerMediumPause and Ponder Q3
You pedal hard along a flat road, then stop pedalling and brake to a halt at a red light. Where does the energy from your muscles go at each stage?
Show solution
Pedalling: chemical energy from your food becomes kinetic energy of you and the bicycle. Part of it warms your body, and part is used against air resistance and friction in the tyres and chain, ending up as heat and a little sound.
Braking: the brakes do negative work on the bicycle. Its kinetic energy becomes heat in the brake pads and rims, the tyres and the road.

Quick check

1 of 3

A toy car at rest gains

3 J

of energy when you push it along a smooth floor. How much work did your push do on it?

0 J

1.5 J

3 J

6 J

Show answer

Answer (C) $3 J$

By the work–energy theorem the work done equals the gain in energy,

3 J .

No factor of ½ appears in the theorem itself.

7.3 Forms of Energy

So far our energy has belonged to moving or raised objects: mechanical energy. But energy turns up in many other forms, and it keeps changing from one form to another. Table 7.1 lists the main forms with everyday examples.

Table 7.1Forms of energyNCERT Fig. 7.10 · p. 121

Form	Where it is found	Everyday example
Mechanical	in moving objects and raised or squeezed ones	a moving bus; water held behind a dam
Chemical	locked in the way atoms are joined in food and fuels	dal and rice, LPG, a torch cell
Electrical	in charges driven round a circuit	the current running a ceiling fan
Thermal	in the jiggling particles of a substance; more when hotter	a hot iron
Light	travels as radiation, even through empty space	sunlight falling on a solar panel
Sound	travels as vibrations through air, water or solids	the beat of a dhol
Nuclear	released when the centres of atoms split or join	the Sun; a nuclear power station

Devices are energy changers. A torch turns the chemical energy in its cells into electrical energy, and then into light. A mixer-grinder turns electrical energy into the kinetic energy of its spinning blades. A solar water heater on a rooftop turns light into thermal energy. You are an energy changer too: climbing the stairs, you turn chemical energy from food into potential energy. Nearly every change also gives off some heat, which is why a phone charger feels warm.

4/4In a home, a lamp turns the electrical energy into light, plus a little heat.

Fig. 7.5One energy chain, from water held behind a dam to light in a home.

A reservoir of water held behind a dam sits on high ground at the left. A pipe runs steeply down to a small power house with a turbine, and a wire runs on to a house with a lamp. In step one a tag reads potential, by the stored water. In step two water flows down the pipe and a second tag reads kinetic. In step three the turbine has turned and a third tag reads electrical. In step four the lamp glows and a fourth tag reads light plus heat.

From here on we focus on mechanical energy, because it connects directly to the forces and motion you met in Chapters 4 and 6.

Quick check

1 of 2

Match each device with the main energy change it makes.

Tap an item on the left, then its match on the right.

Show answer

Answer

solar cell : light → electrical
electric iron : electrical → thermal
torch cell : chemical → electrical
loudspeaker : electrical → sound

Each device takes in one form of energy and gives out another; some heat is almost always produced too.

7.4 Mechanical Energy

An object's mechanical energy (यांत्रिक ऊर्जा) comes from how it moves and where it is. It has two parts: kinetic energy, from motion, and potential energy, from position or shape. To put a number on each part, we ask how much work it takes to give an object that energy.

7.4.1 Kinetic energy

A thrown stone can crack a window pane; a speeding bus can crumple a car. Anything moving carries kinetic energy (गतिज ऊर्जा), the energy of motion. An object at rest has none, so a moving object's kinetic energy is exactly the work it took to get it from rest up to its present speed (Fig. 7.6).

3/3After a distance $s$ its speed is $v .$ The work $F \times s$ has become extra kinetic energy.

Fig. 7.6Work done by a steady push shows up as extra kinetic energy.NCERT Fig. 7.11 · p. 121

A cart of mass m stands on a level track. In the first step it moves slowly with speed u, shown by a short arrow. In the second step a force arrow F pushes it from behind and it has moved part of the way, a little faster. In the third step it has moved a distance s, marked by a dimension line, and a longer arrow shows its new speed v. A label reads: work done W = F × s, all of it becomes extra kinetic energy.

DerivationWork done to change an object's speed

A constant force $F$ acts on an object of mass $m$ along its motion. Its speed rises from $u$ to $v$ while it moves a distance $s .$ A constant force gives a constant acceleration $a,$ so the equations of motion from Chapter 4 apply. Start from $v^{2} = u^{2} + 2 a s$ and solve it for $s :$

s = \frac{v^{2} - u^{2}}{2 a}

(7.4)

The work done is $W = F \times s,$ and Newton's second law gives $F = m a .$ Substitute both:

W = m a \times \frac{v^{2} - u^{2}}{2 a}

W = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

(7.5)

The acceleration cancels. The answer depends only on the starting and final speeds, not on how hard or how gently we pushed. By the work–energy theorem this work is the change in the object's energy. If it starts from rest ( $u = 0$ ), the work done is its kinetic energy at speed $v .$

Kinetic energy

K = \frac{1}{2} m v^{2}

(7.6)

where

$K$: kinetic energy (J)
$m$: mass (kg)
$v$: speed (m/s)

Three quick rules follow from (7.5). If the total work done on an object is positive, it speeds up and gains kinetic energy; if negative, it slows down and loses some; if zero, its speed does not change. Because $v$ is squared, kinetic energy climbs steeply with speed: double the speed and the kinetic energy becomes four times as large, which is why a crash at $80 k m / h$ is far worse than one at $40 k m / h .$ Like work, kinetic energy is a scalar measured in joules.

Example 7.4NCERT Example 7.4 · p. 122

Slowing down in a school zone

A scooter slows from $60 k m / h$ to $20 k m / h$ as it enters a school zone. (a) What fraction of its kinetic energy is left? (b) A second scooter has twice the mass and also moves at $20 k m / h .$ How does its kinetic energy compare with the first scooter's?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.5NCERT Example 7.5 · p. 122

Kinetic energy of a javelin

A javelin of mass $800 g$ leaves an athlete's hand at $108 k m / h .$ Find its kinetic energy.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.6NCERT Example 7.6 · p. 123

Was the car speeding?

A car of mass $1200 k g$ brakes hard on a level road and stops in $30 m .$ Tyre marks show that the braking force was about $4500 N$ throughout. The speed limit is $50 k m / h .$ Was the driver within the limit?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Pause and ponder: questions with answers

NumericalMediumPause and Ponder Q4
A car of mass $1000 k g$ and a loaded truck of mass $9000 k g$ have the same kinetic energy. Which is moving faster, and how many times faster?
Show solution
AnswerThe car, three times as fast.
$\frac{1}{2} (1000) v_{car}^{2} = \frac{1}{2} (9000) v_{truck}^{2}$
$\frac{v_{car}^{2}}{v_{truck}^{2}} = 9$
Taking square roots, $v_{car} = 3 v_{truck} .$ The lighter vehicle must move faster to carry the same kinetic energy.
Quick answerEasyPause and Ponder Q5
An express train running at a steady $80 k m / h$ has more kinetic energy as it passes the second station than as it passed the first.
TrueFalse
Show answer
Answer False
Kinetic energy depends only on mass and speed, $K = \frac{1}{2} m v^{2},$ not on where the train is. Same speed, same kinetic energy.

Quick check

1 of 3

A football of mass

0.45 k g

flies at

20 m / s .

What is its kinetic energy?

Show answer

Answer 90 J

K = \frac{1}{2} \times 0.45 \times 20^{2} = 90 J .

7.4.2 Potential energy

Energy can also sit in storage, waiting to be used. Wind up a toy car and set it down: it scoots off on its own. Push a dart into a toy dart gun and a spring is squeezed; pull the trigger and the spring flings the dart out. A pole vaulter bends her pole on the way up, and as it straightens it lifts her over the bar. In each case someone first did work to change the shape of something, and the work was banked in that new shape.

Why does a squeezed spring fling a toy (Fig. 7.7)? Inside the spring, forces between its coils resist being pushed closer together. Squeezing means working against those forces, and the work is not lost: it waits in the spring's new shape. Release it, and the same internal forces push on the toy until the spring is back to its natural length, handing over the stored energy as kinetic energy.

A group of objects that pull or push on one another can also store energy in its arrangement. Lift a school bag off the floor and you are separating the bag and the Earth against their mutual pull; let go and they come together again. Prise a north pole away from a south pole and you work against their attraction; release them and they snap back hard enough to pinch a finger. Charges that attract or repel each other behave in the same way.

Definition

Potential energy

Stored energy that depends on how a system is arranged: the shape of a deformed object, such as a squeezed spring, or the positions of objects that attract or repel, such as a raised ball and the Earth.

ExampleWater in a rooftop tank has potential energy because it is high above the taps.

1/2Work has been done on each system, and the energy is stored.

Fig. 7.7Energy stored by a change of shape (spring) or of positions (magnets, ball and Earth).NCERT Figs. 7.14–7.16 · p. 124

Three panels side by side, labelled spring, magnets, and ball plus Earth. In the first step a spring is squeezed between a wall and a block, a pair of bar magnets with a south pole facing a north pole is held apart, and a ball is held above the ground at a dashed height line. In the second step the spring has stretched out and pushed the block away, the magnets have moved together and touch, and the ball has dropped to the ground, each with an arrow showing the motion. Labels at the bottom read shape, positions and positions.

Gravitational potential energy

In a ball–Earth pair, nearly all the moving is done by the ball: the Earth is unimaginably heavier and hardly shifts. That is why people simply say “the ball's potential energy” (स्थितिज ऊर्जा). Energy stored this way, by height, is gravitational potential energy. It is the only kind of potential energy we calculate in the rest of this chapter, always with $g = 10 m / s^{2} .$

Activity 7.1NCERT Activity 7.1 · p. 125

Dents in a tray of flour

You’ll need

a deep tray or baking tin
flour or fine dry sand, about 5 cm deep
a heavy steel ball or a large stone marble
a metre scale and a small ruler

Predict first

Which drop will leave the deepest dent?

from 25 cmfrom 50 cmfrom 100 cmall will be equally deep

Smooth the flour so that its surface is flat.
Hold the ball $25 c m$ above the surface and let it drop. Lift it out carefully.
Repeat from $50 c m$ and from $100 c m,$ each time over a fresh, flat patch of flour.
Stand a ruler in each dent to measure its depth, and fill in the table.

Table 7.2Dents in the flour

Drop height	Your prediction (rank 1–3)	Depth of dent (cm)
25 cm
50 cm
100 cm

Your entries are saved automatically.

DerivationPotential energy of a raised object

Take the ground as the level where potential energy is zero. To lift an object of mass $m$ slowly to a height $h,$ you must push up with a force equal to its weight, $m g .$ The force acts along the displacement, so the work you do is force × distance, $m g \times h :$

W = m g h

(7.7)

By the work–energy theorem, this work is stored as the object's potential energy at height $h .$

Gravitational potential energy

U = m g h

(7.8)

where

$U$: potential energy (J)
$m$: mass (kg)
$g$: acceleration due to gravity, $10 m / s^{2}$ in this chapter
$h$: height above the chosen zero level (m)

Its unit is once again the joule. Notice that $U$ depends only on how high the object finishes, not on the route it took to get there.

Example 7.7NCERT Example 7.7 · p. 126

A bag on the loft

A $5 k g$ bag of rice is lifted from the floor onto a loft $2.4 m$ high. How much potential energy does it gain? Take $g = 10 m / s^{2} .$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 1 question

Pause and ponder: questions with answers

Short answerEasyPause and Ponder Q6
A delivery drone flies level at a steady speed, $20 m$ above the ground. Later it rises slowly and steadily to $35 m .$ How does its potential energy change during each part of the flight?
Show solution
Level flight: its height stays $20 m,$ so its potential energy does not change, however far it flies.
Rising: its height grows by $15 m,$ so its potential energy increases by $m g \times 15 m .$

Quick check

1 of 3

2 k g

flowerpot sits on a shelf

1.5 m

above the floor. Taking the floor as zero, what is its potential energy? (

g = 10 m / s^{2}

)

Show answer

Answer 30 J

U = m g h = 2 \times 10 \times 1.5 = 30 J .

7.4.3 Conservation of mechanical energy

Add the two kinds together and you get the mechanical energy, $E = K + U .$ What happens to it when an object falls freely? Let us keep the books on a stone dropped from rest, ignoring air resistance.

DerivationEnergy of a falling object

At the start, point A, an object of mass $m$ is at rest at height $h .$

\begin{aligned} K_{A} & = 0 \\ U_{A} & = m g h \\ E_{A} & = m g h \end{aligned}

(7.9)

After falling for a time $t$ it reaches point B. Its acceleration is $g,$ so its speed is $v = g t$ and it has fallen a distance $d = \frac{1}{2} g t^{2} .$ Its height is now $h - d,$ and its kinetic energy is $\frac{1}{2} m v^{2} = \frac{1}{2} m g^{2} t^{2} = m g d .$ So at B the energies, and their total $E_{B} = U_{B} + K_{B},$ are

\begin{aligned} U_{B} & = m g h - m g d \\ K_{B} & = m g d \\ E_{B} & = m g h \end{aligned}

(7.10)

The kinetic energy gained, $m g d,$ is exactly the potential energy lost. The total stays $m g h$ at every instant, all the way down (Fig. 7.8).

2/3At $10 m$ : $100 J$ potential and $100 J$ kinetic.

Fig. 7.8As the ball falls, potential energy turns into kinetic energy; the total stays the same.NCERT Fig. 7.19 · p. 126

On the left a 1 kg ball falls from point A at 20 m, past point B at 10 m, to point C at ground level. On the right three bars show potential energy, kinetic energy and the total. In the first step the ball is at A at rest: potential 200 J, kinetic 0 J. In the second step it is at B moving at about 14 m/s: potential 100 J and kinetic 100 J. In the third step it is at C moving at 20 m/s: potential 0 J and kinetic 200 J. The total bar stays at 200 J throughout.

LawConservation of mechanical energy

When gravity is the only force doing work on an object, its mechanical energy $K + U$ stays constant. Potential energy lost is gained as kinetic energy, and kinetic energy lost is gained as potential energy.

Activity 7.2NCERT Activity 7.2 · p. 127

Does the bob climb back?

You’ll need

a heavy nut, key or small stone as the bob
about 1 m of strong thread
a nail or hook high on a wall or door frame
a sheet of chart paper and tape
a pencil and a ruler

Predict first

On the far side, the bob will rise

well above the linealmost up to the lineonly halfway to the line

Tie the bob to the thread and hang it from the nail so that it can swing close to the wall without touching it.
Tape the chart paper on the wall behind the bob. Pull the bob to one side and draw a level line on the paper through its centre.
Release the bob from rest at the line; do not push it. Watch where it turns round on the far side.
Note how high it rises on the far side on the 1st, 5th and 10th swings.

Table 7.3How high does it climb?

Swing	On the far side, the bob turns round…
1st
5th
10th

Your entries are saved automatically.

2/4Moving fastest: all kinetic energy.

Fig. 7.9Potential and kinetic energy trade places at every swing.NCERT Fig. 7.20 · p. 127

A pendulum hangs from a support, with a dashed release level drawn through the bob's starting position on the left. Two bars at the right show potential and kinetic energy. In step one the bob is held at the left end on the release level: the potential bar is full and the kinetic bar empty. In step two the bob is at the lowest point: potential empty, kinetic full. In step three it is at the right end, back up on the release level: potential full again. In step four, many swings later, it turns round below the release level and the potential bar is only about half full, with the label lower each swing.

Conservation of mechanical energy turns some hard problems into a few lines of working. Let us answer the water-park question from the start of the chapter.

Example 7.8NCERT Example 7.8 · p. 128

Speed at the bottom of a water slide

A water slide starts $3.2 m$ above the pool. A rider sets off from rest at the top, and water makes the slide almost frictionless. (a) How fast is the rider moving at the bottom? (b) Would a rider of twice the mass, or a slide of a different shape from the same height, change the answer? Take $g = 10 m / s^{2} .$

Try it yourself first. Then reveal the solution one step at a time.

The shape still decides how long the ride takes: a slide that drops steeply at first gets you moving sooner. Test this, and the effect of friction, in the lab below.

InteractiveWater-slide energy lab

A side view of a slide starting from a tower, with a rider at the top. Buttons choose a straight, curved or bumpy slide, all from the same height; sliders set the height from 2 to 8 m and the rider's mass from 20 to 80 kg; two buttons switch friction off (wet) or on (dry). Pressing Play slides the rider down while four bars show potential energy, kinetic energy, heat and the total, and readouts show the speed and the time. A message at the bottom gives the final speed: without friction it is the square root of 2gh for every shape and mass, and with friction it is lower, with the energy turned into heat listed.

Fig. 7.10Pick a slide shape, a height and a rider's mass, then press Play and watch the energy bars.

Example 7.9NCERT Example 7.9 · p. 128

Braking run on a roller coaster

A roller-coaster train of mass $400 k g$ (with its riders) enters an uphill braking run at $12 m / s .$ Along the run the track rises $1 m$ for every $4 m$ travelled along it, and brake fins push back on the train with a steady $1000 N .$ How far along the run does the train go before it stops? Take $g = 10 m / s^{2} .$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Pause and ponder: questions with answers

NumericalMediumPause and Ponder Q7
A $0.3 k g$ guava falls from a branch $5 m$ above the ground. Ignoring air resistance, find its potential, kinetic and total mechanical energy at the branch, halfway down, and just before it lands. Take $g = 10 m / s^{2} .$
Show solution
At the branch, $U = m g h = 0.3 \times 10 \times 5 = 15 J$ and $K = 0.$ Halfway down the height is $2.5 m,$ so $U = 7.5 J$ and the other $7.5 J$ is kinetic. Just before landing $U = 0$ and $K = 15 J,$ a speed of $10 m / s .$
Table 7.4
PE (J) KE (J) Total (J)
At the branch 15 0 15
Halfway down 7.5 7.5 15
Just before landing 0 15 15
The total stays $15 J$ (that is, $m g h$ ) all the way down.
Short answerMediumPause and Ponder Q8
In a science park, a ball is released from rest at the top of a track with a row of humps. Each hump is lower than the one before it. Describe the energy changes as the ball rolls, and explain why the humps have to get lower.
Show solution
Rolling down a slope, potential energy turns into kinetic energy; climbing the next hump, kinetic energy turns back into potential energy, so the ball is slowest at the top of each hump. Friction and air resistance keep turning a little mechanical energy into heat, so the ball's total keeps falling. It can never climb as high as it started, so every hump must be lower than the last, or the ball would roll back.

	PE (J)	KE (J)	Total (J)
At the branch	15	0	15
Halfway down	7.5	7.5	15
Just before landing	0	15	15

Quick check

1 of 4

A stone is dropped from a height of

20 m .

Ignoring air resistance, how fast is it moving just before it hits the ground? (

g = 10 m / s^{2}

)

m/s

Show answer

Answer 20 m/s

\frac{1}{2} m v^{2} = m g h,

v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 20} = 20 m / s .

7.5 Power

Two water pumps fill identical rooftop tanks from the same underground sump. One fills its tank in 10 minutes; the other takes 40 minutes. Both lift the same mass of water through the same height, so both do the same work. The faster pump is not doing more work; it is doing the work more quickly. The rate of doing work is called power (शक्ति).

Average power

P = \frac{W}{t}

(7.11)

where

$P$: average power (watts, W)
$W$: work done (J)
$t$: time taken (s)

Power is measured in watts (W). A watt means one joule of work done every second, $1 W = 1 J / s .$ Bigger powers are given in kilowatts, $1 k W = 1000 W .$ More work in the same time, or the same work in less time, both mean more power (Fig. 7.11). Take care not to mix up the unit W (watt) with the symbol $W$ for work.

3/3B takes $60 s :$ $P = 3000 / 60 = 50 W .$ Same work, a quarter of the power.

Fig. 7.11Same work, different times, different power.

Two identical staircases, A and B, each rise 6 m. A label says: same climb, 3000 J each. In the first step both climbers stand at the bottom. In the second step climber A has reached the top, with the labels 15 s and 200 W, while climber B is only on the first step. In the third step B has also reached the top, with the labels 60 s and 50 W.

Example 7.10NCERT Example 7.10 · p. 129

Filling a rooftop tank

A pump lifts $500 k g$ of water from a sump to a tank $12 m$ higher in 4 minutes. What is its power output? Take $g = 10 m / s^{2} .$

Try it yourself first. Then reveal the solution one step at a time.

Example 7.11NCERT Example 7.11 · p. 129

Power of an e-rickshaw

An e-rickshaw with its passengers has a mass of $600 k g .$ It speeds up from rest to $36 k m / h$ in $8 s$ on a level road. Ignoring friction and air resistance, what average power does its motor deliver?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

40 k g

student runs up a staircase

6 m

high in

12 s .

What power does she develop in lifting herself? (

g = 10 m / s^{2}

)

Show answer

Answer 200 W

W = m g h = 40 \times 10 \times 6 = 2400 J,

P = 2400 / 12 = 200 W .

7.6 Simple Machines

Nobody heaves a full gas cylinder straight onto a delivery truck if a plank is handy, and nobody drags a bucket out of a deep well hand over hand if there is a wheel to run the rope over. Such tools do not shrink the job: the cylinder still gains the same potential energy. What they change is the force you supply, making it smaller or pointing it in a more comfortable direction. Devices like these are simple machines (सरल मशीन). This section looks at three: pulleys, inclined planes and levers.

Two words help. The effort is the force you put into a machine. The load is the force the machine has to beat, often the weight being raised. Their ratio tells you how much the machine multiplies your force:

Mechanical advantage

MA = \frac{load}{effort}

(7.12)

Definition

Mechanical advantage

The ratio of the load a machine overcomes to the effort applied to it.

ExampleA lever with an MA of 5 lets a $100 N$ push lift a $500 N$ load.

An MA greater than 1 means the effort is smaller than the load. The price is that the effort must move through a longer distance, so the work you put in is never less than the work done on the load.

7.6.1 Pulley

A pulley (घिरनी) is simply a grooved wheel, free to turn on an axle, with a rope running in its groove. Over a village well, one hangs from a beam above the opening: you haul down on one end of the rope, and the bucket tied to the other end comes up.

A fixed pulley does not reduce the force you need. To raise a $200 N$ bucket you still pull with about $200 N$ (slightly more, because of friction), so its mechanical advantage is 1. What it gives you is a better direction. Pulling down is easier and safer than heaving upwards, because you can lean back and use your own weight (Fig. 7.12).

2/3Still $200 N,$ so MA = 1, but now you pull down, helped by your own weight.

Fig. 7.12A fixed pulley changes the direction of the effort; a movable pulley also reduces its size.NCERT Figs. 7.24, 7.25 · p. 130

Three scenes, one per step. First, a person on a balcony pulls a bucket straight up on a rope: an upward effort arrow of 200 N and a downward load arrow of 200 N, labelled no machine. Second, a pulley wheel hangs from a beam; the rope runs over it from the bucket down to a person on the ground who pulls downwards: effort 200 N, load 200 N, MA = 1. Third, the bucket hangs from a movable pulley; the rope runs from the beam down round the movable pulley, up over a fixed pulley and down to the person: effort 100 N, load 200 N, MA = 2, with the note that two ropes hold the load.

Quick check

1 of 2

An ideal single fixed pulley is used to raise a

300 N

load. The effort needed is about

300 N

150 N

100 N

600 N

Show answer

Answer (A) $300 N$

A fixed pulley only changes the direction of the force: effort = load, so MA = 1. Halving the effort needs a movable pulley.

7.6.2 Inclined plane

Watch a shopkeeper load a heavy drum of oil onto a truck. He seldom lifts it straight up. He leans a couple of planks against the truck and pushes or rolls the drum up the slope. A sloping surface used this way is an inclined plane (आनत तल). How much does the slope actually help?

Activity 7.3NCERT Activity 7.3 · p. 131

Up the ramp

You’ll need

a toy truck or small cart with free-running wheels
a spring balance
a stack of books about 30 cm high
two smooth boards, about 60 cm and 100 cm long

Predict first

Compared with lifting the truck straight up, pulling it up the longer board will need

more forceless forcethe same force

Hook the spring balance to the truck. Lift the truck slowly and steadily straight up to the top of the books, and note the reading.
Lean the $60 c m$ board against the top of the stack. Pull the truck slowly up the board, keeping the balance parallel to the board, and note the reading.
Repeat with the $100 c m$ board, which makes a gentler slope.
For each method, multiply the force by the distance the truck moved.

Table 7.5Lifting versus pulling up a slope

Method	Distance moved (cm)	Force (N)	Force × distance
Lift straight up	30
60 cm board	60
100 cm board	100

Your entries are saved automatically.

DerivationMechanical advantage of an inclined plane

Push a load of weight $m g$ steadily up a smooth inclined plane of length $L$ onto a platform at height $h$ (Fig. 7.13). Let the effort, applied along the slope, be $F^{'} .$ Ignoring friction, the work you do equals the potential energy the load gains:

F^{'} \times L = m g h

So $F^{'} = m g h / L,$ and the mechanical advantage is

MA = \frac{m g}{F^{'}} = \frac{L}{h}

(7.13)

The plank is always longer than the height it climbs, so $L / h$ is more than 1 and the effort is less than the load. A longer, gentler ramp to the same height gives a bigger MA and an even smaller effort.

2/3MA = 2: $150 N$ along $2 m .$

Fig. 7.13A longer ramp needs less force over a longer distance; the work stays the same.NCERT Fig. 7.28 · p. 132

A platform 1 m high stands at the right; a label reads load 300 N. In the first step a box is lifted straight up beside the platform with an upward force arrow, and the text reads F = 300 N, distance 1 m, F × d = 300 J. In the second step a 2 m plank leads up to the platform with the box on it and a force arrow along the slope: F = 150 N, distance 2 m, F × d = 300 J. In the third step a longer 3 m plank makes a gentler slope: F = 100 N, distance 3 m, F × d = 300 J.

Example 7.12NCERT Example 7.12 · p. 132

A ramp into a shop

A shop's entrance has a step $25 c m$ high. To wheel a loaded handcart in, the shopkeeper lays a straight ramp that starts $60 c m$ in front of the step, measured along the ground. (a) How long is the ramp? (b) What is its mechanical advantage? (c) Ignoring friction, what force along the ramp would push a $520 N$ handcart up it?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Pause and ponder: questions with answers

Short answerMediumPause and Ponder Q9
Ghat roads up to hill stations climb in long zigzags with hairpin bends instead of going straight up the slope. Use the idea of an inclined plane to explain why.
Show solution
A zigzag road is a long, gentle inclined plane. For the same rise $h$ the length $L$ is much greater, so $L / h,$ the MA, is larger and a vehicle needs a much smaller driving force to climb. It travels further, but its engine can manage the gentle slope, and its brakes can control it on the way down.
Short answerMediumPause and Ponder Q10
A worker can reach a rooftop either by climbing a vertical ladder or by walking up a long sloping ramp. Why does the ramp feel easier, even though the gain in potential energy is the same?
Show solution
Either way the gain in potential energy is $m g h .$ On the vertical ladder all of it is supplied over the short distance $h,$ so at each step the worker must lift the full body weight. On the ramp the same energy is spread over a much longer distance, so the force needed at each step is smaller.

Quick check

1 of 3

A smooth plank

3 m

long leads up to a platform

0.6 m

high. What is its mechanical advantage?

Show answer

Answer 5

MA = L / h = 3 / 0.6 = 5.

7.6.3 Lever

Activity 7.4NCERT Activity 7.4 · p. 133

A ruler that lifts

You’ll need

a 30 cm ruler
a round marker pen or thick round pencil
a small steel bowl (katori)
a handful of identical coins

Predict first

Where should the pen go so that the fewest coins lift the bowl?

close to the bowlnear the middle of the rulerclose to the coins

Lay the ruler across the pen so that the pen is $5 c m$ from one end. Put the bowl on that short end.
Place coins one at a time on the far end until the bowl lifts. Count them.
Move the pen to $8 c m$ and then to $11 c m$ from the bowl's end, and count again each time.

Table 7.6Coins needed to lift the bowl

Pen placed	Coins needed
5 cm from the bowl's end
8 cm from the bowl's end
11 cm from the bowl's end

Your entries are saved automatically.

Your ruler was a lever (उत्तोलक): a stiff bar that pivots on a support. The pivot is the fulcrum (आलंब). The load presses on the bar at one place and your effort at another, and what matters is how far each one is from the fulcrum. We call these distances the load arm and the effort arm (Fig. 7.14).

Tilt the bar and compare its two ends. In Fig. 7.14 your end sweeps down through a big distance $d_{1}$ while the load end creeps up through a small distance $d_{2} .$ The bar itself neither stores nor makes energy, so (ignoring friction) the work you put in with effort $F_{1}$ equals the work done on the load by the force $F_{2}$ :

F_{1} \times d_{1} = F_{2} \times d_{2}

(7.14)

With $d_{1}$ several times $d_{2},$ the push $F_{2}$ on the load is several times your effort $F_{1} .$ A lever trades distance for force: it cuts the force you need, never the work you do.

1/2Fulcrum, load and effort, with the load arm and the effort arm.

Fig. 7.14A long effort arm lets a small effort lift a large load.NCERT Fig. 7.32 · p. 133

A long iron rod rests on a triangular fulcrum near its left end. A heavy chest labelled load sits on the short left end, and a downward effort arrow acts at the far right end. Dimension lines below mark the short load arm, from the load to the fulcrum, and the long effort arm, from the fulcrum to the effort. In the second step the rod has tilted: the right end has moved down a large distance d1 and the chest has risen a small distance d2, and a dashed outline shows where the rod was.

Activity 7.5NCERT Activity 7.5 · p. 133

A metre-scale balance

You’ll need

a metre scale (or any long, light, straight stick)
strong thread
two identical paper cups with thread loops
about 15 identical coins or metal washers
a hook, or a stand to hang the scale from

Predict first

3 coins hang 40 cm from the middle on the left. Where should a cup of 6 coins hang on the right to balance them?

40 cm from the middle30 cm from the middle20 cm from the middle80 cm from the middle

Tie the thread tightly round the scale at its $50 c m$ mark and hang it so that it swings freely. If it does not hang level, slide the thread a little until it does. This point is the fulcrum.
Hang a cup on each side with a thread loop that can slide along the scale.
Put 3 coins in the left cup (the effort) and hang it $40 c m$ from the fulcrum. Keep it there.
Put 3, then 4, then 6, then 8 coins in the right cup (the load). Each time, slide the right cup until the scale is level, and measure its distance from the fulcrum.

Table 7.7Balancing the metre scale (effort: 3 coins at 40 cm, 3 × 40 = 120)

Load coins	Balancing distance (cm)	Coins × distance
3
4
6
8

Your entries are saved automatically.

Balanced lever

\begin{aligned} effort \times effort arm \\ = load \times load arm \end{aligned}

(7.15)

Rearranging gives the mechanical advantage of a lever:

\begin{aligned} MA & = \frac{load}{effort} \\ = \frac{effort arm}{load arm} \end{aligned}

(7.16)

A long effort arm and a short load arm give a large MA. As with every machine there is a price: your end moves further than the load does, so the work you do is not reduced.

InteractiveMetre-scale balance

A metre scale hangs from a string at its middle, with a pan on each side. For each pan, plus and minus buttons set the number of identical coins from 1 to 8, and a slider sets its distance from the fulcrum from 5 to 45 cm. The beam tilts towards the side with the larger product of coins and distance and is level when the two products are equal. A message shows both products and, when the beam balances with more coins on the load side, the mechanical advantage as a ratio of coins and of distances. A New challenge button fixes the left pan and asks you to slide the right pan to the balancing position.

Fig. 7.15Add coins and slide the pans until the beam balances, then try a challenge.

Example 7.13NCERT Example 7.13 · p. 134

Balancing a seesaw

A park seesaw has its fulcrum in the middle and seats every $0.5 m$ along each arm, up to $2 m$ from the fulcrum. Meera ( $24 k g$ ) sits $1.5 m$ from the fulcrum. (a) Where should Kabir ( $36 k g$ ) sit on the other side to balance her? (b) Meera's little brother ( $12 k g$ ) climbs onto her seat with her. Where must Kabir move now?

Try it yourself first. Then reveal the solution one step at a time.

Classes of levers

Levers are sorted into three classes by what sits in the middle: the fulcrum, the load or the effort. Fig. 7.16 and Table 7.8 show all three.

1/3Fulcrum in the middle, as in a seesaw or a pair of scissors.

Fig. 7.16What sits in the middle decides the class.NCERT Table 7.2 · p. 135

The same horizontal bar is shown three times. In step one, Class I, the fulcrum is under the middle, the load sits on the left end and a downward effort arrow acts at the right end; examples: seesaw, scissors, crowbar. In step two, Class II, the fulcrum is at the left end, the load is in the middle and an upward effort arrow acts at the right end; examples: wheelbarrow, nutcracker, bottle opener. In step three, Class III, the fulcrum is at the left end, an upward effort arrow acts in the middle and the load is at the right end; examples: tweezers, tongs, broom.

Table 7.8Classes of leversNCERT Table 7.2 · p. 135

Class	In the middle	Mechanical advantage	Examples
I	fulcrum	more or less than 1	seesaw, scissors, crowbar, hand-pump handle
II	load	always more than 1	wheelbarrow, nutcracker, bottle opener
III	effort	always less than 1	tweezers, fishing rod, your forearm

A Class III lever needs more effort than the load. It is still useful, because the load end moves further and faster than your hand: a broom sweeps a wide arc, and tweezers give fine control.

Look closely at everyday machines and you will find simple machines inside: a village hand pump works its piston with a lever, and a bicycle's brake handles are levers too. No machine is a source of energy. At best it passes on every joule you give it; real ones always lose a few joules to friction as heat.

Practice · 3 questions

Pause and ponder: questions with answers

Short answerMediumPause and Ponder Q11
Why is a bottle opener with a long handle easier to use than one with a very short handle?
Show solution
The opener is a lever. Its tip rests on top of the cap (the fulcrum), and its lip hooks under the edge of the cap (the load), only a centimetre or so away. A longer handle gives a longer effort arm while the load arm stays the same, so the MA (effort arm ÷ load arm) is larger and a smaller push lifts the cap.
Short answerMediumPause and Ponder Q12
A gardener cutting a thick twig with pruning shears pushes it deep into the jaws, close to the pivot. Why?
Show solution
The shears are a pair of levers with the pivot as fulcrum. Placing the twig near the pivot makes the load arm short while the effort arm (the handles) stays long. The MA becomes large, so the blades press on the twig with a much bigger force than the gardener's grip.
Short answerHardPause and Ponder Q13
An online video shows a wheel with swinging weights that seems to turn on its own forever and even lights a bulb, with no battery, fuel or push. Use the ideas of work and energy to explain why no such perpetual motion machine can work.
Show solution
Friction at the axle and air resistance do negative work on the wheel all the time, turning its mechanical energy into heat, and lighting a bulb takes energy out as well. Nothing supplies energy to replace these losses, and energy cannot be created. So the wheel's kinetic energy must run down until it stops. A machine that seems to run forever must have a hidden energy source.

Quick check

1 of 4

A crowbar has an effort arm of

1.2 m

and a load arm of

0.15 m .

What is its mechanical advantage?

Show answer

Answer 8

MA = 1.2 / 0.15 = 8.

Recap

Work, Energy, and Simple Machines: key points and summary

A force does work on an object when the object moves along the force: $W = F \times s,$ measured in joules (Section 7.1).
Work is zero if there is no displacement, no force along the motion, or a force at right angles to the motion (Section 7.1.1).
Work is positive when force and displacement point the same way and negative when they are opposite (Section 7.1.2).
Energy is the capacity to do work. Work–energy theorem: the work done on an object equals the change in its energy (Section 7.2).
Energy takes many forms (mechanical, chemical, electrical, thermal, light, sound, nuclear) and changes from one form to another (Section 7.3).
Kinetic energy is energy of motion, $K = \frac{1}{2} m v^{2}$ (Section 7.4.1).
Potential energy is energy stored by shape or position; near the Earth, $U = m g h$ (Section 7.4.2).
When only gravity does work, mechanical energy $K + U$ is conserved; friction turns some of it into heat (Section 7.4.3).
Power is the rate of doing work, $P = W / t,$ measured in watts, $1 W = 1 J / s$ (Section 7.5).
Simple machines change the size or direction of a force; mechanical advantage = load ÷ effort (Section 7.6).
A fixed pulley has an MA of 1 but lets you pull down (Section 7.6.1); an inclined plane has MA $= L / h$ (Section 7.6.2).
A lever balances when effort × effort arm = load × load arm, so its MA = effort arm ÷ load arm (Section 7.6.3).
No machine reduces the total work needed or creates energy.

Flash cards · 7

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In your NCERT textbook	Practice question here
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Quick check: MCQs on Work, Energy, and Simple Machines

7.1.2 Positive and negative work done4 questions
7.2 The Work-Energy Theorem3 questions
7.3 Forms of Energy2 questions
7.4.1 Kinetic energy3 questions
Gravitational potential energy3 questions
7.4.3 Conservation of mechanical energy4 questions
7.5 Power3 questions
7.6.1 Pulley2 questions
7.6.2 Inclined plane3 questions
Classes of levers4 questions

Glossary · 15

Key terms (English–Hindi)

energyऊर्जाoorja: The capacity to do work, measured in joules.First seen in the chapter →

fulcrumआलंबaalamb: The fixed point about which a lever turns.First seen in the chapter →

inclined planeआनत तलaanat tal: A sloping surface used to raise a load with a smaller force over a longer distance.First seen in the chapter →

jouleजूलjool: The SI unit of work and energy: the work done when a $1 N$ force moves an object $1 m$ along it.First seen in the chapter →

kinetic energyगतिज ऊर्जाgatij oorja: Energy an object has because it is moving, $K = \frac{1}{2} m v^{2} .$ First seen in the chapter →

leverउत्तोलकuttolak: A rigid bar that turns about a fixed point, the fulcrum.First seen in the chapter →

mechanical advantageयांत्रिक लाभyaantrik laabh: Load divided by effort for a machine.
mechanical energyयांत्रिक ऊर्जाyaantrik oorja: The sum of an object's kinetic and potential energy.First seen in the chapter →

potential energyस्थितिज ऊर्जाsthitij oorja: Energy stored because of shape or relative position; near the Earth, $U = m g h .$ First seen in the chapter →
powerशक्तिshakti: The rate of doing work, $P = W / t .$ First seen in the chapter →
pulleyघिरनीghirni: A grooved wheel that guides a rope; a fixed pulley changes the direction of the effort.First seen in the chapter →

simple machineसरल मशीनsaral masheen: A device such as a pulley, ramp or lever that changes the size or direction of a force to make a task easier.First seen in the chapter →

wattवाटvaat: The SI unit of power: one joule per second.First seen in the chapter →
workकार्यkaarya: Done by a force when the object it acts on moves along the force; for a constant force, $W = F \times s .$ First seen in the chapter →
work-energy theoremकार्य-ऊर्जा प्रमेयkaarya-oorja pramey: The work done on an object equals the change in its energy.

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

What you will learn in Work, Energy, and Simple Machines

7.1 Work Done by a Constant Force

7.1.1 When is work done equal to zero?

7.1.2 Positive and negative work done

Pause and ponder: questions with answers

7.2 The Work-Energy Theorem

Pause and ponder: questions with answers

7.3 Forms of Energy

7.4 Mechanical Energy

7.4.1 Kinetic energy

Pause and ponder: questions with answers

7.4.2 Potential energy

Gravitational potential energy

Pause and ponder: questions with answers

7.4.3 Conservation of mechanical energy

Pause and ponder: questions with answers

7.5 Power

7.6 Simple Machines

7.6.1 Pulley

7.6.2 Inclined plane

Pause and ponder: questions with answers

7.6.3 Lever

Classes of levers

Pause and ponder: questions with answers

Work, Energy, and Simple Machines: key points and summary

Practice questions with answers (NCERT “Revise, reflect, refine”)

Find your NCERT exercise question

Quick check: MCQs on Work, Energy, and Simple Machines

Key terms (English–Hindi)

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