class 9 science chapter 7 notes and book solutions

Class 9 Science — Chapter 7

Work, Energy, and Simple Machines

Short Concept Notes and NCERT Book Solutions

Part 1: Short Concept Notes

1. Work Done by a Force

In science, work is done when a force causes displacement in the direction of that force.

Work done = Force × Displacement
W = F × s

SI unit: joule (J).
One joule of work is done when a force of 1 N moves an object 1 m in the direction of the force.
Positive work: force and displacement are in the same direction.
Negative work: force acts opposite to displacement.
Zero work: displacement is zero, force is zero, or force is perpendicular to displacement.

Example: Pushing a trolley forward does positive work on it. Friction acting backward does negative work.

Remember: Feeling tired does not always mean scientific work has been done on an object. If the wall does not move, work done on the wall is zero.

2. Work–Energy Theorem

Work done on an object equals the change in its energy.

Work done = Change in energy

When positive work is done, an object may gain energy. When negative work is done, it may lose energy. Energy can be transferred from one object to another during a collision.

3. Forms of Energy

Energy is the capacity to do work. Common forms include:

Mechanical energy: energy due to motion or position.
Thermal energy: associated with temperature and particle motion.
Light energy: allows us to see.
Sound energy: carried by vibrations and sound waves.
Electrical energy: associated with electric charges.
Chemical energy: stored in food, fuels and other substances.
Nuclear energy: stored in atomic nuclei.

Examples of energy conversion: an electric bulb changes electrical energy into light and heat; food provides chemical energy to our muscles.

4. Kinetic Energy

Kinetic energy is the energy an object has because it is moving.

Kinetic energy = ½mv²

Here, m is mass in kilograms and v is speed in metres per second.

A stationary object has zero kinetic energy.
For the same speed, a heavier object has more kinetic energy.
Kinetic energy depends on the square of speed.

If speed doubles, kinetic energy becomes four times as large because (2v)² = 4v².

Example: For a 2 kg object moving at 3 m/s, KE = ½ × 2 × 3² = 9 J.

5. Potential Energy

Potential energy is stored energy due to an object’s position, shape or configuration.

Gravitational potential energy

PE = mgh

Here, m is mass, g is acceleration due to gravity, and h is height.

A 2 kg object raised 3 m, using g = 10 m/s², gains PE = 2 × 10 × 3 = 60 J.

Elastic potential energy

A stretched rubber band, compressed spring or bent bow can store energy due to deformation.

Near Earth’s surface, gravitational potential energy depends on mass, gravity and height. For the same object at the same final height, the gain in potential energy does not depend on the path taken.

6. Conservation of Mechanical Energy

Mechanical energy is the sum of kinetic energy and potential energy.

Mechanical energy = KE + PE

When only conservative forces such as gravity act and friction is negligible, the total mechanical energy remains constant.

As an object falls, its potential energy decreases and its kinetic energy increases.
At the highest point of an upward throw, its instantaneous speed is zero, so its kinetic energy is zero and its gravitational potential energy is greatest.
Friction and air resistance transfer some mechanical energy into heat and sound.

Educational ramp model showing a cart and inclined plane

Related picture: a ramp model used to explore motion and energy.

7. Power

Power is the rate at which work is done.

Power = Work ÷ Time
P = W/t

SI unit: watt (W).
1 watt = 1 joule per second.
Doing the same work in less time requires greater power.
1 horsepower is approximately 746 W.

If 600 J of work is done in 3 s, power = 600 ÷ 3 = 200 W.

8. Simple Machines and Mechanical Advantage

Simple machines make tasks easier by changing the magnitude or direction of the force applied. They do not create energy.

Mechanical advantage (MA) = Load ÷ Effort

Pulley

A pulley is a grooved wheel used with a rope. A fixed pulley changes the direction of effort but, ideally, has mechanical advantage 1. Simple pulley lifting a weight

Related picture: a simple pulley system.

Inclined plane

A ramp helps raise a load using a smaller force over a longer distance.

Ideal MA of an inclined plane = Length ÷ Height
MA = L/h

Adjustable inclined plane used for physics demonstrations

Related picture: an adjustable inclined plane.

Lever

A lever is a rigid bar that rotates about a fixed point called the fulcrum.

Fulcrum: pivot point.
Load: force or object to be moved.
Effort: force applied to the lever.

Effort × Effort arm = Load × Load arm

Ideal MA of lever = Effort arm ÷ Load arm

Lever diagram showing effort, load and fulcrum

Related picture: parts of a lever.

Lever class	Position	Examples
Class I	Fulcrum between effort and load	Seesaw, scissors
Class II	Load between fulcrum and effort	Wheelbarrow, bottle opener
Class III	Effort between fulcrum and load	Tweezers, broom

A machine can reduce the effort force, but the effort usually has to act over a larger distance. In real machines, friction also means that some input energy becomes heat.

Part 2: NCERT Textbook Questions and Solutions

Question 1

State whether True or False.(i) Work is said to be done when a force is applied, even if the object does not move.(ii) Lifting a bucket vertically upward results in positive work done on the bucket.(iii) The SI unit for both work and energy is joule (J).(iv) A motionless stretched rubber band has kinetic energy.(v) Energy can change from one form to another.

Answers

(i) False. Work on an object requires displacement in the direction of the force.

(ii) True. The applied force and the bucket’s displacement are both upward.

(iii) True. Both work and energy are measured in joules (J).

(iv) False. A stretched rubber band at rest has elastic potential energy, not kinetic energy.

(v) True. Energy can change from one form to another.

Question 2

Fill in the blanks.(i) Work done = ______ × ______ (in the direction of force).(ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.(iii) The expression for kinetic energy of a body of mass m and velocity v is ______.(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______.(v) Power is defined as the ______ at which work is done.

Answers

(i) Force × displacement.

(ii) 1 newton.

(iii) ½mv².

(iv) mgh.

(v) Rate.

Question 3

When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?(i) The force acting on the ball is zero.(ii) The acceleration of the ball is zero.(iii) Its kinetic energy is zero.(iv) Its potential energy is maximum.

Answer

Correct statements: (iii) and (iv).

At the highest point, the ball’s instantaneous velocity is zero, so its kinetic energy is zero. Its height and gravitational potential energy are maximum. Gravity still acts downward, so the force and acceleration are not zero.

Question 4

For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.

Answers

(i) A truck moving uphill: Kinetic energy is converted into gravitational potential energy as the truck gains height.

(ii) Unwinding of a watch spring: Elastic potential energy is converted into mechanical energy and motion.

(iii) Photosynthesis in green leaves: Light energy is converted into chemical energy stored in food.

(iv) Water flowing from a dam: Gravitational potential energy is converted into kinetic energy.

(v) Burning of a matchstick: Chemical energy is converted into heat and light energy.

(vi) Explosion of a fire cracker: Chemical energy is converted into heat, light, sound and kinetic energy.

(vii) Speaking into a microphone: Sound energy is converted into electrical signals.

(viii) A glowing electric bulb: Electrical energy is converted into light and heat energy.

(ix) A solar panel: Light energy is converted into electrical energy.

Question 5

A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s–2, and student’s mass is m = 50 kg.(i) Find the gain in the potential energy if the student is lifted straight up to the top.(ii) Find the gain in the potential energy when the student climbs the stairs to the same top.(iii) What do you conclude about the dependence of the potential energy on the path taken?

Solution

(i) Lifted by elevator

Gain in PE = mgh

= 50 × 10 × 72.5 = 36,250 J

(ii) Climbing the stairs

The student reaches the same height, so the gain is also 36,250 J.

(iii) Conclusion

Gain in gravitational potential energy depends on the change in height, not on the path taken.

Answers: (i) 36,250 J; (ii) 36,250 J; (iii) Potential energy gain does not depend on the path taken.

Question 6

A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.

Solution

Let the height of one floor be h and the first time be t.

Energy required to lift to the 10th floor:

E₁ = mg(10h) = 10mgh

Energy required to lift to the 20th floor:

E₂ = mg(20h) = 20mgh = 2E₁

The second lift requires twice as much energy.

First power:

P₁ = 10mgh/t

Second power, taking twice the time:

P₂ = 20mgh/(2t) = 10mgh/t = P₁

Answer: The second lift requires twice the energy, but the same average power.

Question 7

Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.

Answer

The energy required depends mainly on the mass of the flag, the height raised and gravitational acceleration. Ideally, the work is mgh.

Raising the flag slowly or quickly does not change the ideal work done if the same flag is raised through the same height. It changes the time taken.

Power = work ÷ time. If the same work is completed in half the time, power doubles.

Answer: Energy depends on mass and height. Doubling the raising speed ideally doubles the power required, if the lifting speed and time change proportionally and other conditions remain the same.

Question 8

A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.

Solution

Fuel energy supplied is proportional to the increase in kinetic energy.

First day: Total mass = 100 + 60 = 160 kg.

KE₁ = ½ × 160 × v² = 80v²

Second day: Total mass = 100 + 60 + 40 = 200 kg.

KE₂ = ½ × 200 × v² = 100v²

Fuel ratio = KE₁ : KE₂ = 80 : 100 = 4 : 5

Answer: First day : second day = 4 : 5.

Question 9

On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.

Answer

For balance, clockwise moment = anticlockwise moment.

Child’s weight × child’s distance
= Adult’s weight × adult’s distance

Let the child’s weight be W and the adult’s weight be 2W.

W × d₁ = 2W × d₂
d₁ = 2d₂

The child must sit twice as far from the fulcrum as the adult.

Answer: The child sits at twice the distance from the fulcrum compared with the adult.

Question 10

A ball of mass 2 kg is thrown up with a velocity of 20 m s–1.(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s–2).

Solution

(i) Upward motion: Gravity acts downward while the ball moves upward. Work done by gravity is negative.

Downward motion: Gravity and the ball’s motion are both downward. Work done by gravity is positive.

(ii) Work done by air resistance

Initial KE = ½mv²
= ½ × 2 × 20² = 400 J

PE at 19.4 m = mgh
= 2 × 10 × 19.4 = 388 J

At the highest point, the ball’s kinetic energy is zero. The loss of mechanical energy is due to air resistance.

Work by air resistance = 388 − 400 = −12 J

Answer: (i) Negative during upward motion and positive during downward motion. (ii) Work done by air resistance = −12 J.

Question 11

A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?

Solution

Given: Mass = 10.0 kg; initial kinetic energy = 180 J.

(i) Speed at 0 m

KE = ½mv²

180 = ½ × 10 × v²

v² = 36, so v = 6 m/s

(ii) Speed at 4 m

Work done is the area under the force–displacement graph.

From 0 to 1 m: triangle area = ½ × 1 × 50 = 25 J.
From 1 to 3 m: rectangle area = 2 × 50 = 100 J.
From 3 to 4 m: triangle area = ½ × 1 × 50 = 25 J.

Total work = 25 + 100 + 25 = 150 J

Final KE = Initial KE + Work
= 180 + 150 = 330 J

330 = ½ × 10 × v²

v² = 66
v = √66 ≈ 8.12 m/s

Negative acceleration: No. The force shown in the graph is in the direction of motion throughout the 0–4 m interval, so the acceleration is not negative.

Answers: (i) 6 m/s; (ii) approximately 8.12 m/s; no negative acceleration.

Reference: Tiwari Academy — Chapter 7 solutions

Question 12

The gravitational attraction on the surface of the Moon (lunar surface) is about 1 th 6 of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?

Solution

For the same initial upward velocity, maximum height is inversely proportional to gravitational acceleration.

Height on Moon = 6 × Height on Earth

= 6 × 8 = 48 m

Answer: The ball will reach a height of 48 m on the Moon.

Question 13

A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38.(i) Describe how the car moves between positions A and B.(ii) Calculate the kinetic energy of the car at A.(iii) State the work done by the brakes in bringing the car to a halt between B and C.(iv) What does the kinetic energy of the car transform into?

Solution

(i) Between A and B, the car moves at a constant speed of 35 m/s, as shown by the horizontal part of the speed–time graph.

(ii) Kinetic energy at A

KE = ½mv²

= ½ × 1000 × 35²

= 500 × 1225 = 612,500 J

(iii) Work done by the brakes

The car comes to rest, so its final kinetic energy is zero.

Work done = Change in KE
= Final KE − Initial KE

= 0 − 612,500 = −612,500 J

(iv) The kinetic energy is mainly converted into heat due to friction in the brakes and between the tyres and road. Some energy is also converted into sound.

Answers: (i) Constant speed of 35 m/s; (ii) 612,500 J; (iii) −612,500 J; (iv) Mainly heat energy, with some sound energy.

Reference: Learn Insta — Chapter 7 solutions

Question 14

The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s–1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Solution

Mass of the ball = 0.5 kg. At O, velocity is zero and potential energy is 30 J. Therefore, total mechanical energy is 30 J. As the track is frictionless, total mechanical energy remains constant.

At P: Potential energy from the graph = 20 J.

KE = 30 − 20 = 10 J

KE = ½mv²

10 = ½ × 0.5 × v²

v² = 40, so v = √40 ≈ 6.32 m/s

At Q: Potential energy from the graph = 30 J.

KE = 30 − 30 = 0 J

v = 0 m/s

At R: The potential energy shown by the graph is 10 J.

KE = 30 − 10 = 20 J

20 = ½ × 0.5 × v²

v² = 80, so v = √80 ≈ 8.94 m/s

Answers: Velocity at P ≈ 6.32 m/s; at Q = 0 m/s; at R ≈ 8.94 m/s.

Reference: Learn Insta — Chapter 7 solutions

Question 15

A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.(i) Calculate the velocity of the coconut just before it hits the sand.(ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s–2.

Solution

Given: m = 1.5 kg, h = 10 m, g = 10 m/s², resistive force = 3000 N.

(i) Velocity before impact

v² = u² + 2gh

The coconut starts from rest, so u = 0.

v² = 2 × 10 × 10 = 200

v = √200 ≈ 14.14 m/s

(ii) Depth of depression

Potential energy lost by the coconut is used to do work against the sand’s resistive force.

Energy = mgh = 1.5 × 10 × 10 = 150 J

Work = Force × depth

150 = 3000 × depth

Depth = 150 ÷ 3000 = 0.05 m

0.05 m = 5 cm

Answers: (i) Approximately 14.14 m/s; (ii) Depth of depression = 0.05 m or 5 cm.