Chapter 7 - Work, Energy and Simple Machines

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

Strengthen your understanding of Class 9 Science Chapter 7: Work, Energy and Simple Machines with chapter-wise NCERT Solutions designed for effective learning and exam preparation. Aligned with the new NCERT Exploration textbook (2026–27) and CBSE guidelines, these expert-created solutions feature step-by-step answers to every in-text and end-of-chapter question, along with important questions for focused practice. Curated by ALLEN Experts, the solutions are available for free PDF download, making revision and offline study convenient anytime, anywhere.

1.0Download NCERT Solutions for Class 9 Science Chapter 7 – Work, Energy and Simple Machines PDF

Class 9 Science Chapter 7: Work, Energy and Simple Machines builds on the concepts of force and motion introduced in Chapter 6 and explains how forces are described through work, energy, and power. Students learn that work is done when a force causes displacement, explore the Work-Energy theorem, understand kinetic and potential energy, the law of conservation of energy, and how simple machines like pulleys, inclined planes, and levers make everyday tasks easier. Students can download the free PDF of NCERT Solutions for Class 9 Science Chapter 7: Work, Energy and Simple Machines, featuring step-by-step solutions to all textbook questions, prepared according to the latest NCERT syllabus and CBSE-aligned guidelines.

NCERT Solutions Class 9 Science Chapter 7

2.0Learning Outcomes of NCERT Class 9 Science Chapter 7 Solutions

Understand the Scientific Concept of Work: Define work as done when a force causes displacement, and calculate work using W = F × d × cos θ.
Identify Conditions for Zero Work: Recognise situations where work done is zero, such as when force and displacement are perpendicular or displacement is zero.
Apply the Work-Energy Theorem: Understand that the net work done on an object equals the change in its kinetic energy.
Define and Calculate Kinetic Energy: Derive and apply the formula KE = ½mv² to solve numerical problems.
Define and Calculate Potential Energy: Understand gravitational potential energy (PE = mgh) and energy due to position or deformation.
Understand Mechanical Energy: Explain mechanical energy as the sum of kinetic and potential energy.
State the Law of Conservation of Energy: Explain how mechanical energy is conserved in the absence of non-conservative forces like friction and air resistance.
Define Power and Calculate It: Understand power as the rate of doing work and calculate it using P = W/t, along with its SI unit, the watt.
Understand Simple Machines: Explain how pulleys, inclined planes, and levers make work easier by changing the force required.
Calculate Mechanical Advantage: Determine the mechanical advantage of simple machines and understand the trade-off between force and distance.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports future topics in mechanics, energy conservation, and higher-level physics.

3.0Detailed NCERT Class 9 Science Chapter 7 Work, Energy and Simple Machines Solutions

1. What will be the magnitude of velocity of the child at the bottom of the blue slide? Will two children of different masses reach the bottom of the same slide with the same velocity? Which of the slides will result in the largest magnitude of velocity for the child at its bottom?

Solution

The child's gravitational potential energy at the top changes into kinetic energy at the bottom. The magnitude of velocity at the bottom depends mainly on the vertical height of the slide (assuming negligible friction). Two children of different masses will reach the bottom with the same velocity if they start from the same height because mass cancels out in the calculation. The slide with the greatest vertical drop (height difference) will give the child the largest velocity at the bottom.

2. In the previous chapter, a weightlifter is shown holding a barbell steady in her hands. Is she doing any work on the barbell while holding it steady?

Fig. 6.8: A weightlifter lifting barbell

Solution

No, she is not doing any mechanical work on the barbell while holding it steady.

Reason: Work $=$ Force × Displacement Since the barbell does not move, its displacement is zero, so the work done on the barbell is zero.

3. Is the work done by friction on the stack of coins that travels on a rough surface (positive, negative or zero)?

Solution

The work done by friction is negative. Reason: Friction acts opposite to the direction of motion.

A force opposite to displacement does negative work and removes kinetic energy from the moving coins.

4. When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?

Solution

The energy supplied by your muscles appears as:

Kinetic energy of the bicycle and rider. Thermal energy (heat) due to friction in the bicycle parts and with the road.

5. Two objects A and B of mass m and 4 m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B ?

Solution

Given Masses $= m$ and $4 m$ Kinetic energies are equal. Using $KE = \frac{1}{2} mv^{2}$ $\frac{1}{2} m v_{A}^{2} = \frac{1}{2} (4 m) v_{B}^{2}$ $v_{A}^{2} - 4 v_{B}^{2}$ $v_{A} = 2 v_{B}$ Therefore $v_{A} : v_{B} = 2 : 1$

6. Does the kinetic energy of an object which moves with constant velocity change with its position?

Solution

No. Since $KE = \frac{1}{2} mv^{2}$ and the velocity remains constant, the kinetic energy remains constant and does not depend on position.

7. Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?

Solution

(a) Horizontal motion with constant velocity: Height does not change. Therefore, gravitational potential energy remains unchanged. (b) Gradually raised vertically:

Height increases. Since PE $= mgh$ potential energy increases as the object is raised.

8. For the situation depicted in figure calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.

Solution

At the highest point, the ball has: Potential Energy $=$ mgh Kinetic Energy = 0 So, total mechanical energy $=$ mgh. Just before the ball hits the ground: Height $h = 0$ , so Potential Energy $= 0$ The ball's velocity is $v$ , so Kinetic Energy $= \frac{1}{2} mv^{2}$ Using $v^{2} = 2 gh$ , $KE = \frac{1}{2} m (2 gh) = mgh$ Therefore, the mechanical energy just before the ball hits the ground is $ME = KE + PE = mgh + 0 = mgh$ The mechanical energy just before the ball hits the ground is mgh. When the ball is released from the highest point: At the top of a hill (A, C, E): Potential Energy is maximum. Kinetic Energy is minimum. At the bottom of a valley ( $B, D$ ): Potential Energy is minimum. Kinetic Energy is maximum.

9. You may have seen an exhibit like that in figure in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points $A, B$ and C . Why do subsequent points, such as $C, D$ and E , usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Why do the heights of C, D and E decrease?

Solution

As the ball moves, some of its mechanical energy is lost due to:

Friction between the ball and track Air resistance This lost energy is converted into heat and sound. Therefore, the ball cannot climb back to the same height each time, so successive peaks ( $C, D, E$ ) are lower than the previous ones.

Yes. The decreasing heights are due to energy lost because of friction and air resistance.

10. (i) Work is said to be done when a force is applied, even if the object does not move.

Solution: False

In physics, primary condition of work to be done requires displacement. The formula is Work = Force x Displacement. In case displacement is zero, work done is also zero. (ii) Lifting a bucket vertically upward results in positive work done on the bucket.

Solution True

In physics, work is done only when there is displacement. The formula for work is: $W = F \times d$ If the displacement is zero, then the work done is also zero. (iii) The SI unit for both work and energy is joule (J).

Solution True

Energy is the ability to do work. Both energy and work have the same physical dimensions and are measured in the same standard units.

(iv) A motionless stretched rubber band has kinetic energy.

Solution

False

Kinetic energy is the energy possessed by a moving object. If an object is at rest, its velocity is zero, so it has no kinetic energy; however, it may store elastic potential energy instead. (v) Energy can change from one form to another.

Solution True

According to the law of conservation of energy, energy can neither be created nor destroyed; it can only change from one form to another, such as electrical energy changing into light energy.

11. Fill in the blanks: (i) Work done $=$ Force × displacement (in the direction of force). (ii) 1 joule of work is done when a force of 1 (one) 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 $\frac{1}{2} mv^{2}$ . (iv) The potential energy of an object of mass $m$ at a small height $h$ from the Earth's surface is mgh. (v) Power is defined as the rate at which work is done.

12. 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.

Solution

Correct statements: (iii) and (iv) Explanation: Force is not zero (gravity acts downward) Acceleration is not zero ( g acts downward)

Velocity becomes zero $= KE = 0$ Height is maximum $= PE$ is maximum

13. 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 firecracker (vii) Speaking into a microphone (viii) A glowing electric bulb (ix) A solar panel

Solution

(i) A truck moving uphill

When a truck moves uphill, it gains height. Its kinetic energy is gradually converted into gravitational potential energy as it comes to stop.

Energy transformation: Kinetic energy to Potential energy (ii) Unwinding of a watch spring

A wound spring stores elastic potential energy. When it unwinds, this stored energy is used to produce motion in the hands of watch.

Energy transformation: Potential energy to Kinetic energy (iii) Photosynthesis in green leaves Plants use sunlight to prepare food during photosynthesis. Light energy is converted into stored chemical energy in plants.

Energy transformation: Light energy to Chemical energy (iv) Water flowing from a dam Water stored at height has potential energy. As it flows downward, this energy changes into energy of motion.

Energy transformation: Potential energy to Kinetic energy (v) Burning of a matchstick

The chemical energy stored in the matchstick is released as heat and light when it burns. Energy transformation: Chemical energy to Heat energy + Light energy (vi) Explosion of a firecracker Firecrackers contain chemical energy, which is suddenly released as heat, light, sound, and kinetic energy. Energy transformation: Chemical energy $=$ Heat + Light + Sound + Kinetic energy (vii) Speaking into a microphone

When a person speaks, sound energy is produced and converted into electrical signals by the microphone. Energy transformation: Sound energy to Electrical energy (viii) A glowing electric bulb

Electrical energy supplied to the bulb is converted into light and heat.

Energy transformation: Electrical energy to Light energy + Heat energy (ix) A solar panel

Solar panels convert sunlight directly into electrical energy. Energy transformation: Light energy to Electrical energy.

14. 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 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) The gain in gravitational potential energy is given by: $PE = mgh$ Substituting the values: PE $= 50 \times 10 \times 72.5$ $= 50 \times 725$ $= 36250 J$ The gain in potential energy $= 36, 250 J$ (ii) The gain in potential energy will be the same as in part (i), because the height reached is the same. So, Gain in potential energy = 36,250 J Potential energy depends only on mass, gravity, and height, not on how the height is reached. (iii) The potential energy does not depend on the path taken; it depends only on the initial and final positions (vertical height). Whether the student goes straight up in an elevator or climbs the stairs, the gain in potential energy remains the same because both reach the same height.

15. A crane lifts a mass $m$ to the 10 th 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 height of each floor $= h$ Height to 10th floor $= 10 h$ Height to 20th floor $= 20 h$ Energy $=$ mgh For 10th floor: $E_{1} := mg (10 h) = 10 mgh$ For 20th floor: $E_{2} = mg (20 h) = 20 mgh$ So, $E_{2} = 2 E_{1}$ Energy required is double. Power = Work / Time Let time for 10th floor $= t$ Time for 20th floor $= 2 t$ $P_{1} = E_{1} : / t = 10 mgh / t$ $P_{2} = E_{2} /2 t = 20 mgh /2 t = 10 mgh / t$ So, $P_{2} = P_{1}$ Power required remains the same. Energy required is doubled. Power required remains the same.

16. 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.

Solution

Factors determining the energy required: The energy required to raise the flag is equal to the gain in its gravitational potential energy.

Potential energy (PE) = mgh Therefore, the factors are: Mass of the flag (m) Height of the flagpole (h) Acceleration due to gravity (g) Effect of speed on work done: Raising the flag slowly or quickly does NOT change the amount of work done. Work done depends only on force and displacement ( $W =$ mgh in this case), not on time or speed. Since the flag is raised to the same height, the work done remains the same in both cases.

Effect on power when speed is doubled: Power is defined as: Power = Work / Time

If the speed of raising the flag is doubled, the time taken becomes half.

So, New power $=$ Work $/ ($ Time $/2) = 2 \times$ (Work/Time) Therefore, power becomes double. Conclusion: Energy depends on mass, height, and gravity.

Work done remains the same whether the flag is raised slowly or quickly.

If speed is doubled, the power required also doubles.

17. 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 consumption is proportional to energy required

Energy required = Kinetic Energy (KE) according to speed $KE = \frac{1}{2} mv^{2}$ Since the final velocity is the same on both days, KE depends only on the total mass.

Day 1: Mass of man $= 60 kg$ , Mass of scooter $=$ 100 kg , Total mass $= 60 + 100 = 160 kg$

Kinetic energy: KE1: $= \frac{1}{2} \times 160 \times v^{2}$ Day 2: Mass of man $= 60 kg$ , Mass of son $= 40 kg$ , Mass of scooter $= 100 kg$ Total mass $= 60 + 40 + 100 = 200 kg$ Kinetic energy: $KE_{2} = \frac{1}{2} \times 200 \times v^{2}$ Ratio of fuel used: Fuel consumption is proportional to KE $KE_{1} : KE_{2} = 160 : 200 = 4 : 5$ The ratio of fuel used on the two days is 4 : 5.

18. 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: Work done by gravity is negative (force opposite to motion)

Downward motion: Work done by gravity is positive (force in direction of motion) (ii) Initial $KE = 12 mv^{2} = \frac{1}{2} \times 2 \times (20)^{2} = 400 J$

Potential energy at height:

$PE = m g h = 2 \times 10 \times 19.4 = 388 J$

Work done by air resistance = Change in mechanical energy $=$ Final energy - Initial energy $= 388 - 400 = - 12 J$ Work done by air resistance $= - 12 J$

19. A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in figure, 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 kg Initial KE = 180 J (i) Speed at 0 m :

$KE = \frac{1}{2} mv^{2} = 180 = 12 \times 10 \times v^{2} v^{2} = 36, v = 6 m / s$

Work done $=$ Area under forcedisplacement graph

From graph:

$0 - 1 m = triangle = \frac{1}{2} \times 1 \times 50 = 25 J 1 - 3 m = rectangle = 2 \times 50 = 100 J 3 - 4 m = triangle = \frac{1}{2} \times 1 \times 50 = 25 J Total work = 25 + 100 + 25 = 150 J Final KE = Initial KE + Work = 180 + 150 = 330 J$

(ii) Speed at 4 m :

$330 = \frac{1}{2} \times 10 \times v^{2}, v^{2} = 66, v = 66 = 8.12 m / s$

Negative acceleration: No, because the force is always in the direction of motion.

20. The gravitational attraction on the surface of the Moon is about 1/6th 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 have thrown with the same upward velocity travel from the surface of the Moon?

Solution

The maximum height reached by a ball thrown upward depends on the acceleration due to gravity. Using the relation: $h = u^{2} / (2 g)$ For the same initial velocity ( $u$ ), height is inversely proportional to g : h proportional to $1/ g$ Given: Height on Earth $= 8 m$ Acceleration due to gravity on Moon $= (\frac{1}{6}) \times g$ (Earth) Since gravity on the Moon is $1/6$ th of that on Earth, the height reached will be 6 times greater. Height on Moon: $h_{m} = 6 \times 8$ $h_{m} = 48 m$ The ball will rise to a height of 48 m on the surface of the Moon. Due to lower gravitational pull on the Moon, the ball experiences less downward acceleration, allowing it to rise to a greater height for the same initial velocity.

21. 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 figure. (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 with a constant speed of $35 m s^{- 1}$ . This is because the speed-time graph is a horizontal straight line in this interval, which shows that the speed does not change with time. Therefore, the car is in uniform motion and no acceleration acts on it during this part of the motion. (ii) Given:

Mass of the car, $m = 1000 kg$ Speed at A, v = $35 m s^{- 1}$ Kinetic Energy $= \frac{1}{2} mv^{2}$ $= \frac{1}{2} \times 1000 \times (35)^{2}$ $= 500 \times 1225$ $= 612500 J$ The kinetic energy of the car at A is $6, 12, 500 J$ . (iii) When the brakes are applied, the car slows down and finally stops at C . The work done by the brakes is equal to the change in kinetic energy of the car. Initial kinetic energy at $B = 6, 12, 500 J$ Final kinetic energy at $C = 0 J$ Work done by brakes $=$ Final $KE -$ Initial KE = 0-612500 $= - 612500 J$ The work done by the brakes is $6, 12, 500 J$ . The negative sign shows that the braking force acts opposite to the direction of motion of the car. (iv) The kinetic energy of the car is mainly transformed into heat energy due to friction between the brake pads and the wheels and between the tyres and the road. A small part may also be converted into sound energy.

22. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. At 0, 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, $m = 0.5 kg$ , At point O: Velocity $= 0 m s^{- 1}$ Potential energy = 30 J Since the track is frictionless, the total mechanical energy of the ball remains constant.

At O: Kinetic energy at $0 = \frac{1}{2} mv^{2} = 0$ So, total mechanical energy $=$ Potential energy + Kinetic energy $= 30 + 0 = 30 J$ Therefore, at every point on the track: Potential energy + Kinetic energy = 30 J From the graph: Potential energy at $P = 20 J$ Potential energy at Q = 30 J Potential energy at R=40 J Now we calculate the velocity at each point.

At P: Kinetic energy at P = Total energy Potential energy at P $30 - 20 = 10 J$ Using, $KE = \frac{1}{2} mv^{2}$ $10 = \frac{1}{2} \times 0.5 \times v^{2}$ $10 = 0.25 v^{2}$ $= v^{2} = 40, v = 40$ $v = 6.32 m s^{- 1}$

So, velocity at $P = 6.32 m s^{- 1}$ At Q: Kinetic energy at $Q = 30 - 30 = 0 J$ Thus, $v = 0 m s^{- 1}$ So, velocity at $Q = 0 m s^{- 1}$ At R: Potential energy at R = 40 J , which is greater than the total mechanical energy (30 J).

This is not possible for the ball because its total energy remains constant at 30J.

Hence, the ball cannot reach point $R$ . So, velocity at R cannot be calculated because the ball never reaches $R$ .

Velocity at $P = 6.32 m s^{- 1}$ Velocity at Q $= 0 m s^{- 1}$ The ball cannot reach $R$ .

23. 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 the coconut's energy is used to create depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume $g = 10 m s^{- 2}$ .

Solution

(i) Given:

Mass of coconut, $m = 1.5 kg$ Height, $h = 10 m$ Acceleration due to gravity, $g = 10 m s^{- 2}$ Initial velocity, $u = 0$ Using the equation: $v^{2} = u^{2} + 2 gh$ $v^{2} = 0 + 2 \times 10 \times 10 = 200$ $v = 200 = 102, v = 14.14 ms^{- 1}$ The velocity of the coconut just before hitting the sand is $14.14 m s^{- 1}$ . (ii) Just before striking the sand, the coconut has kinetic energy equal to the loss in potential energy during the fall.

Kinetic energy on impact: $KE = mgh$ $= 1.5 \times 10 \times 10 = 150 J$ This entire energy is used to do work against the resistive force of sand.

Work done $=$ Force x distance So, $150 = 3000 \times d$ $d = \frac{150}{3000}, d = 0.05 m$ , Converting into centimeters: $0.05 m = 5 cm$

4.0Key Topics in NCERT Class 9 Science Chapter 7 Work, Energy and Simple Machines

Topic	What Students Learn
Work Done by a Constant Force	Understand work as the product of force and displacement in the direction of the force.
Conditions for Zero Work	Identify cases where no work is done despite the presence of force or displacement.
Work-Energy Theorem	Relate the net work done on an object to the change in its kinetic energy.
Kinetic Energy	Understand energy possessed by a moving object and derive the formula KE = ½mv².
Potential Energy	Understand energy due to position or configuration, including gravitational potential energy.
Mechanical Energy and Its Conservation	Learn how kinetic and potential energy interconvert while total mechanical energy remains constant.
Power	Understand the rate of doing work and calculate power using P = W/t.
Simple Machines	Explore pulleys, inclined planes, and levers, and how they reduce the effort needed to do work.
Mechanical Advantage	Calculate the ratio of load to effort and understand the effort-distance trade-off in machines.
Applications in Daily Life	Apply concepts of work, energy, and machines to real-life situations like lifting loads, ramps, and levers.

5.0Mind Map / Concept Recap

6.0Related Study Materials Class 9 Science

Study with ALLEN's Class 9 Science study materials in alignment to the latest NCERT syllabus. Revise well Prepare for school and CBSE examinations confidently Access NCERT Solutions NCERT Textbooks Revision Notes Sample Papers Previous Years' Question Papers Strengthen your Concepts

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0 Advantages of Chapter 7 Science Class 9 NCERT Solutions

Explains Nutrition: Helps students understand digestion and the role of nutrients in animals.
Strengthens Respiration Concepts: Simplifies the process of breathing and exchange of gases.
Clarifies Transportation: Explains the functions of the heart, blood, and blood vessels.
Improves Understanding of Excretion: Helps students learn how waste products are removed from the body.
Introduces Coordination: Explains how different body systems work together to maintain life processes.

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

1.0Download NCERT Solutions for Class 9 Science Chapter 7 – Work, Energy and Simple Machines PDF

NCERT Solutions Class 9 Science Chapter 7

2.0Learning Outcomes of NCERT Class 9 Science Chapter 7 Solutions

Understand the Scientific Concept of Work: Define work as done when a force causes displacement, and calculate work using W = F × d × cos θ.
Identify Conditions for Zero Work: Recognise situations where work done is zero, such as when force and displacement are perpendicular or displacement is zero.
Apply the Work-Energy Theorem: Understand that the net work done on an object equals the change in its kinetic energy.
Define and Calculate Kinetic Energy: Derive and apply the formula KE = ½mv² to solve numerical problems.
Define and Calculate Potential Energy: Understand gravitational potential energy (PE = mgh) and energy due to position or deformation.
Understand Mechanical Energy: Explain mechanical energy as the sum of kinetic and potential energy.
State the Law of Conservation of Energy: Explain how mechanical energy is conserved in the absence of non-conservative forces like friction and air resistance.
Define Power and Calculate It: Understand power as the rate of doing work and calculate it using P = W/t, along with its SI unit, the watt.
Understand Simple Machines: Explain how pulleys, inclined planes, and levers make work easier by changing the force required.
Calculate Mechanical Advantage: Determine the mechanical advantage of simple machines and understand the trade-off between force and distance.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports future topics in mechanics, energy conservation, and higher-level physics.