Where Mindful Learning Begins
Where Mindful Learning Begins
Work, Energy and Simple Machines Class 9 NCERT question answers (CBSE 2026-27), with every Think It Over, Activity, Pause & Ponder, Worked Example, Revise Reflect Refine and Journey Beyond question solved step by step. Revise your Class 9 Science work and energy notes first, then use this page to check each numerical.
This chapter explains motion through energy, and these Class 9 Science Chapter 7 Work, Energy and Simple Machines NCERT solutions cover when work is actually done, the work-energy theorem, kinetic and potential energy numericals, conservation of mechanical energy, power, and mechanical advantage and efficiency of levers, pulleys and inclined planes, with every step shown.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
Work, Energy, and Simple Machines builds one powerful idea from the ground up — that the work done on an object equals the change in its energy — and then uses this work-energy theorem to explain kinetic energy, potential energy, conservation of mechanical energy, power, and finally simple machines like the pulley, the inclined plane and the lever. Every question is solved here, section by section, with full working for every numerical, as complete Class 9 Science Chapter 7 NCERT solutions with answers.
What counts as work, when it is zero, positive or negative, and how work done equals the change in an object's energy.
Deriving K = ½mv² and U = mgh, and showing that mechanical energy is conserved in free fall and pendulum motion.
Rate of doing work, and how a pulley, inclined plane, or lever trades off force against distance without reducing total work.
Let the height of the blue slide be \(h\) and the mass of the child be \(m\).
Potential energy of the child at the top of the slide = \(mgh\).
Let the magnitude of velocity at the bottom be \(v\). Kinetic energy at the bottom = \(\tfrac12 mv^2\).
The potential energy at the top gets converted entirely to kinetic energy at the bottom (neglecting friction).
From \(\tfrac12 mv^2 = mgh\), the mass \(m\) cancels on both sides, so \(v = \sqrt{2gh}\).
This velocity depends only upon the height \(h\); the mass of the child does not matter.
The velocity at the bottom is \(v = \sqrt{2gh}\). It depends only upon the height \(h\), and not on the shape of the slide.
In the figure, the blue, red and purple slides all start from the same platform, so they have the same height \(h\).
So the child reaches the bottom of each slide with the same velocity, \(v = \sqrt{2gh}\) (neglecting friction).
If the heights were different, the slide with the greatest height would give the largest velocity.
Step 2: Yes, a depression is created in the sand.
The ball at a height has energy (potential energy). When it falls on the sand, it applies a force on the sand and the energy creates the depression.
Step 3: Yes, there is a difference in the depths of the depressions.
The depression is deepest when the ball is dropped from the greatest height, and shallowest when it is dropped from the smallest height.
Raising a ball to a greater height from the surface of the Earth requires more work. Thus, the ball possesses more energy at greater height.
Observation: Yes, the bob almost reaches the level of the horizontal line on the other side (point R).
At point P, the bob has potential energy equal to \(mgh\), where \(h\) is the height from the bottom-most point. Its kinetic energy is zero.
At point Q, the potential energy is zero but the pendulum has kinetic energy.
At point R, on the other side, the kinetic energy is zero and the bob regains its potential energy.
The bob reaches almost the same height it started with.
Step 3: Yes, the reading of the spring balance is smaller than that of step 2.
Step 4: The force required decreases as the plank becomes less steep.
However, the force has to be applied over a larger distance to bring the cart to the same height.
Observation: A much heavier object (the stapler) is lifted by a much lighter eraser (one eraser, or two erasers).
This is made possible by using the scale as a simple machine called a lever.
A lever is a rigid bar, such as the scale, that can rotate about a fixed point (the fulcrum), such as the point of contact of the scale with the pencil.
Sample readings (1 coin in the left pan at a distance of 20 cm from the fulcrum):
| Number of coins in left pan, n₁ (Effort) | Distance of left pan from the fulcrum, L₁ (cm) | Number of coins in right pan, n₂ (Load) | Distance of right pan from the fulcrum, L₂ (cm) |
|---|---|---|---|
| 1 | 20 | 1 | 20 |
| 1 | 20 | 2 | 10 |
| 1 | 20 | 4 | 5 |
| 1 | 20 | 8 | 2.5 |
In every row, \(n_1 \times L_1 = n_2 \times L_2\) (for example, \(1 \times 20 = 4 \times 5 = 20\)).
No, she is not doing any work on the barbell.
Work done = force × displacement in the direction of the force.
She applies an upward force on the barbell, but the barbell does not move, so the displacement \(s = 0\).
Work done = \(F \times 0 = 0\).
The force of friction acts on the stack of coins in the direction opposite to its motion.
The displacement of the stack of coins is in the forward direction.
So the force and the displacement are in opposite directions.
While pedalling, the chemical energy of the food gives energy to the muscles.
This muscular energy appears as the kinetic energy of the bicycle and the rider.
A part of it appears as thermal energy (heat) due to friction (between the tyres and the road, and in the chain and the wheels).
A small part also appears as sound energy.
Given: mass of A = \(m\), mass of B = \(4m\); kinetic energy of A = kinetic energy of B
To find: \(v_A : v_B\)
Kinetic energy \(K = \tfrac12 mv^2\).
If the object moves with a constant velocity, \(v\) is constant. The mass \(m\) is also constant.
So the kinetic energy is constant, whatever the position of the object.
Potential energy \(U = mgh\), where \(h\) is the height above the ground.
Horizontal motion with constant velocity: the height \(h\) does not change, so the potential energy does not change.
Gradually raised in the vertical direction: the height \(h\) increases, so the potential energy increases.
At point A (height \(h\), initial velocity \(u = 0\)):
potential energy = \(mgh\), kinetic energy = 0, mechanical energy = \(mgh\).
Just before the ball hits the ground (point C), the height is zero, so the potential energy = 0.
Velocity of the ball: \(v^2 = u^2 + 2gh = 2gh\).
Point A (highest point): the ball has maximum potential energy and zero kinetic energy.
Point B (lower point): the potential energy has decreased and the kinetic energy has increased; the kinetic energy is maximum at the lowest point.
Point C: the kinetic energy changes back into potential energy, but C is lower than A.
The subsequent points C, D and E have lower heights because some mechanical energy is lost at each stage.
Yes, it is due to the energy lost because of friction (and air resistance), which converts mechanical energy into heat.
Work done in going up the hill is the same in both cases (\(mgh\)), ignoring friction.
A winding road with a gentle slope has a larger length \(L\) for the same height \(h\).
Mechanical advantage of an inclined plane = \(L/h\), so a larger \(L\) gives a larger mechanical advantage.
So the force (effort) required is smaller, though it has to be applied over a larger distance.
An inclined ladder works like an inclined plane.
Its length \(L\) is larger than the height \(h\), so its mechanical advantage \(L/h\) is greater than 1.
So a smaller force is required to reach the same height, though the distance covered is larger.
The spoon works as a lever; the rim of the can acts as the fulcrum.
The effort is applied at the end of the spoon handle (long effort arm). The load is near the rim (short load arm).
Mechanical advantage = effort arm / load arm, which is greater than 1.
So a small effort gives a large force on the lid.
Effort × effort arm = load × load arm.
When the hard object is pushed closer to the fulcrum, the load arm becomes smaller.
For the same effort, the force applied on the object (load) becomes larger.
(Mechanical advantage = effort arm / load arm becomes larger.)
A machine cannot create energy; it only helps us use it more effectively.
In a real machine, some work is always done against friction.
This work does not get stored but is converted into heat, which is lost.
So the mechanical energy of the machine keeps decreasing, and it slows down and finally stops, unless energy is supplied again.
The girl applies a force equal to the weight of the dumbbell to lift it up.
While moving the dumbbell up: the force is in the direction of displacement, so she does positive work on it.
While moving the dumbbell down: the force she applies to hold it is in the direction opposite to the displacement, so she does negative work on it.
Given: force = 200 N, displacement of the ball in the direction of force = –15 cm = –0.15 m
The goalkeeper applied a force opposite to the motion of the ball, so she did negative work on the ball.
The displacement is taken as negative because the ball moves in the direction opposite to the direction of the applied force.
The moving striker collides with the white coin, which in turn collides with the black coin.
The moving striker applies force in the direction of displacement of the white coin. So the striker does positive work on the white coin, increasing its energy.
By Newton’s third law, the white coin applies an opposite force and does negative work on the striker, decreasing its energy.
Similarly, the white coin does positive work on the black coin, increasing its energy.
The black coin does negative work on the white coin, decreasing its energy.
Let the mass of the vehicle be \(m\) and its initial velocity be \(v\).
Initial kinetic energy = \(\tfrac12 mv^2\).
Kinetic energy at velocity \(2v\):
Given: m = 0.2 kg, \(v = 154.8\text{ km h}^{-1}\)
Velocity of the ball:
Kinetic energy:
Given: m = 15000 kg, force by the wire = 367500 N (backward), distance = 100 m, final velocity = 0
To find: initial velocity \(v\)
Let the aircraft approach with a velocity \(v\).
Initial kinetic energy = \(\tfrac12 \times 15000 \times v^2\). Final kinetic energy = 0 J.
The displacement and the force applied by the wire are in opposite directions, so the work done by the wire is negative:
By the work-energy theorem, work done = change in kinetic energy:
Given: m = 200 g = 0.2 kg, h = 10 m, \(g = 10\text{ m s}^{-2}\)
Potential energy:
Potential energy of the child at the top of the slide = \(mgh\).
Let the magnitude of velocity at the bottom be \(v\). Kinetic energy at the bottom = \(\tfrac12 mv^2\).
The potential energy at the top gets converted entirely to the kinetic energy at the bottom (neglecting friction). Equating the two:
This velocity depends only upon the height \(h\); the shape of the slide and the mass of the child should not matter.
Given: m = 10000 kg, \(v = 72\text{ km h}^{-1} = 20\text{ m s}^{-1}\), force by the sand = 50000 N, \(g = 10\text{ m s}^{-2}\); height gained = d/2
To find: minimum length \(d\) of the ramp
Initial kinetic energy = \(\tfrac12 \times 10000 \times (20)^2 = 2000000\text{ J}\). Initial potential energy = 0 J.
Total initial energy = 2000000 J.
Let the truck travel a distance \(d\) along the ramp. Height gained = \(d/2\).
Final kinetic energy = 0 J. Final potential energy = \(10000 \times 10 \times \tfrac d2 = 50000 \times d\) J.
Change in total energy = \((50000 \times d) - 2000000\) J.
Work done by the sand on the truck = \(-50000 \times d\) J.
Using the work-energy theorem, work done by sand = change in total energy:
Given: m = 75 kg, h = 2 m, t = 5 s, \(g = 10\text{ m s}^{-2}\)
Work done:
Power:
Given: m = 1000 kg, \(u = 0\), \(v = 72\text{ km h}^{-1}\), t = 10 s
Final velocity:
Work done by the engine in 10 s = final kinetic energy – initial kinetic energy:
Power:
Given: height AB = 30 cm, width BC = 40 cm
From Fig. 7.29, the ramp AC is the hypotenuse of the right-angled triangle ABC:
Using Eq. (7.13):
Suppose the child of mass 15 kg sits on seat A and the other child sits at a distance \(L\) from the fulcrum.
Using effort × effort arm = load × load arm (Eq. 7.15):
Work is done only when there is a displacement in the direction of the force.
If the object does not move, \(s = 0\) and the work done is zero.
The force and the displacement are both upward, in the same direction.
The SI unit of work and of energy is the joule (J).
A motionless rubber band has zero velocity, so its kinetic energy is zero.
It has potential energy because it is stretched.
For example, electrical energy changes into light energy in a bulb.
The gravitational force (weight) acts on the ball at every point.
The ball still has the acceleration due to gravity, \(g\), downwards.
At the highest point the velocity is zero, so \(K = \tfrac12 mv^2 = 0\).
The height is maximum, so \(U = mgh\) is maximum.
Given: m = 50 kg, h = 72.5 m, \(g = 10\text{ m s}^{-2}\)
The vertical height gained is the same, h = 72.5 m.
The gain in potential energy is the same in both cases, although the paths are different.
Let the height of each floor be \(H\). Height of the 10th floor = \(10H\); height of the 20th floor = \(20H\).
Energy:
Power: the first task takes time \(t\) and the second takes \(2t\).
Energy required: it depends on the mass of the flag (m), the height of the flagpole (h) and the acceleration due to gravity (g), because \(W = mgh\).
Slowly or quickly: the work done is the same, because \(W = mgh\) does not depend on the time taken.
Power: \(P = W/t\). If the speed is doubled, the time taken is halved.
Given: mass of the scooter = 100 kg; man = 60 kg; son = 40 kg; same final speed \(v\) in the same time
The energy supplied by the fuel appears as the kinetic energy of the scooter with the riders (no other losses).
First day: total mass = 60 + 100 = 160 kg.
Second day: total mass = 60 + 40 + 100 = 200 kg.
Let the mass of the child be \(m\). Then the weight of the adult is twice that of the child, i.e., the mass of the adult is \(2m\).
Let the child sit at a distance \(L_1\) and the adult at a distance \(L_2\) from the fulcrum.
For the seesaw to balance, using Eq. (7.15):
Upward motion: gravity acts downward and the displacement is upward (opposite directions), so the work done by gravity is negative.
Downward motion: gravity and the displacement are both downward (same direction), so the work done by gravity is positive.
Given: m = 2 kg, \(u = 20\text{ m s}^{-1}\), h = 19.4 m, \(g = 10\text{ m s}^{-2}\)
Initial kinetic energy:
At the highest point the kinetic energy is zero and the potential energy is:
By the work-energy theorem, work done by air resistance = change in the total energy:
Given: m = 10.0 kg, kinetic energy at 0 m = 180 J
The work done by the force = area under the force-displacement graph from 0 m to 4 m (a trapezium):
By the work-energy theorem, the kinetic energy at 4 m = 180 + 150 = 330 J.
Negative acceleration? The force is always in the direction of motion (it is never negative), so the acceleration is never negative. The block speeds up throughout.
The ball is thrown up with the same velocity, so it has the same initial kinetic energy \(\tfrac12 mv^2\) on the Earth and on the Moon.
At the highest point this kinetic energy is converted into potential energy \(mgh\):
So \(h \propto 1/g\) for the same \(v\). On the Moon, \(g_M = g/6\), so the height is 6 times:
Given: m = 1000 kg; from the graph, speed at A = 35 m s⁻¹
The speed-time graph is a horizontal line, so the speed stays at 35 m s⁻¹ from 0 s to 1 s.
The car moves with a constant speed (no brakes are applied yet).
The car stops at C, so the final kinetic energy is 0.
By the work-energy theorem, work done = change in kinetic energy:
The kinetic energy changes into heat (the brakes and tyres get hot) and a little sound.
Given: m = 0.5 kg; from the graph, PE at O = 30 J, at P = 20 J, at Q = 30 J, at R = 40 J; velocity at O = 0
The track is frictionless, so the total mechanical energy is conserved. At O: total energy = 0 + 30 J = 30 J.
The potential energy at R (40 J) is more than the total energy (30 J), so the kinetic energy would have to be negative, which is not possible.
Given: m = 1.5 kg, h = 10 m, resistive force of sand = 3000 N, \(g = 10\text{ m s}^{-2}\)
Potential energy at the top is converted into kinetic energy (conservation of mechanical energy):
Energy of the coconut = \(mgh = 1.5 \times 10 \times 10 = 150\text{ J}\).
This energy is used up in the work done against the resistive force of the sand over the depth \(d\):
Energy conversion: the elastic potential energy of the stretched rubber band changes into the kinetic energy of the refill.
Observation: when the rubber band is stretched more, more energy is stored in it.
So the refill gets more kinetic energy and travels a larger distance.
Lever model: a ruler on a pencil (fulcrum). Measure the effort and the load, and the arms.
Inclined plane model: a cardboard ramp. Pull a cart with a spring balance (effort); the weight of the cart is the load.
Mechanical advantage:
Energy Skate Park: watch how kinetic and potential energy change as the skater moves; add friction to see the thermal energy increase.
Energy Forms and Changes: see how energy changes from one form to another.
Pendulum Lab: see the exchange between potential and kinetic energy at different masses, heights and friction.
Masses and Springs: see how energy changes as a mass hangs from a spring.
Work done on an object equals the change in its energy — this single work-energy theorem explains why a moving ball can knock over a wicket, why a raised pot can shatter on impact, why kinetic and potential energy trade places as a pendulum swings while their sum stays constant, and why every simple machine, from a pulley to a lever, can only trade force for distance and never reduce the total work that must be done.
Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.
Given: F = 40 N, s = 5 m
To find: work done \(W\)
Given: m = 0.5 kg, \(h_1 = 20\text{ m}\), \(h_2 = 12.8\text{ m}\), \(g = 10\text{ m s}^{-2}\)
The potential energy is converted into kinetic energy:
The ball rises to 12.8 m, so the kinetic energy just after the rebound equals the potential energy at 12.8 m:
The lost energy is converted into heat and sound (and the deformation of the ball and the ground), so the mechanical energy is not conserved in the collision.
Given: m = 2 kg, h = 5 m, \(g = 10\text{ m s}^{-2}\)
All the potential energy is converted into kinetic energy:
Given: pump A: m = 200 kg, t = 20 s; pump B: m = 300 kg, t = 40 s; h = 10 m; efficiency of A = 80%
Pump A:
Pump B:
Energy consumed by A: the useful work (20000 J) is 80% of the energy supplied:
Given: effort arm = 2 m, load arm = 0.5 m, effort = 100 N
Given: input energy = 500 J, useful output = 350 J
To find: efficiency
At the lowest point, the height of the bob is zero (taking the lowest point as reference), so its potential energy is zero.
The total mechanical energy (PE + KE) stays constant (neglecting friction and air resistance).
So the potential energy lost has been converted into kinetic energy; the bob moves fastest at the lowest point.
On the other side, the kinetic energy changes back into potential energy and the bob rises again.
Revisit Chapter 6 on Forces and Motion, browse the full Class 9 Science hub, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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