NCERT Solutions for Class 9 Science Exploration Chapter 7

Work is done when a force causes displacement in the direction of the force.
Energy is the capacity to do work, while simple machines help change the magnitude or direction of effort without reducing total work.

NCERT Solutions for Class 9 Science Exploration Chapter 7 help students solve Work, Energy, and Simple Machines from the 2026-27 Class 9 Science textbook. The chapter connects force and motion with work, kinetic energy, potential energy, conservation of mechanical energy, power, pulley, inclined plane, lever, and mechanical advantage. Students should focus on the formula-based questions because this chapter uses Physics concepts from earlier chapters on motion and force. These NCERT Class 9 Science Solutions cover Pause and Ponder questions and all Class 9 Science Chapter 7 exercise solutions in a clear step-wise format.

Key Takeaways

• Work: Work depends on force and displacement in the direction of force.
• Energy: Energy is the capacity to do work and is measured in joules.
• Mechanical energy: Kinetic energy and potential energy together form mechanical energy.
• Simple machines: Pulley, inclined plane, and lever make tasks easier by changing effort or direction.

NCERT Solutions for Class 9 Science Exploration Chapter 7 Structure 2026

NCERT Solutions for Class 9 Science Exploration Chapter 7 In-Text Questions

Class 9 Science Exploration Chapter 7 uses examples like lifting bags, saving a football goal, riding a bicycle, falling objects, slides, pulleys, ramps, levers, and seesaws to explain work and energy.

Q1. Is a weightlifter doing work while holding a barbell steady?

Answer: No, the weightlifter is not doing work on the barbell while holding it steady.

Explanation:
In science, work is done only when force causes displacement in the direction of force.

Here, the weightlifter applies an upward force, but the barbell does not move.

So:

$W = F \times s$

Since $s = 0$,

$W = 0$

The weightlifter may feel tired because muscles use energy internally, but no scientific work is done on the barbell.

Q2. Is work done by friction on a moving stack of coins positive, negative, or zero?

Answer: The work done by friction is negative.

Explanation:
Friction acts opposite to the direction of motion. The stack of coins moves forward, while friction acts backward.

When force and displacement are in opposite directions, work done is negative.

So, friction does negative work and reduces the kinetic energy of the stack of coins.

Q3. When you pedal a bicycle on a flat road, in what forms does muscular energy appear?

Answer: Muscular energy appears mainly as kinetic energy of the bicycle and rider, and partly as thermal energy and sound.

Explanation:
When a person pedals, chemical energy from food is converted into muscular energy. This energy helps turn the pedals and move the bicycle.

Some energy becomes kinetic energy of the moving bicycle. Some energy is lost as heat due to friction in tyres, chain, road contact, and air resistance.

A small amount may also become sound energy.

Q4. Two objects A and B of masses $m$ and $4m$ have the same kinetic energy. Find the ratio of their velocities.

Answer: The ratio of velocities of A and B is $2 : 1$.

Explanation:
Kinetic energy is:

$K = \frac{1}{2}mv^2$

For object A:

$K_A = \frac{1}{2}m v_A^2$

For object B:

$K_B = \frac{1}{2}(4m)v_B^2$

Given:

$K_A = K_B$

$\frac{1}{2}m v_A^2 = \frac{1}{2}(4m)v_B^2$

$v_A^2 = 4v_B^2$

$v_A = 2v_B$

So:

$v_A : v_B = 2 : 1$

Q5. Does kinetic energy change if an object moves with constant velocity?

Answer: No, kinetic energy does not change if the object moves with constant velocity.

Explanation:
Kinetic energy depends on mass and velocity.

$K = \frac{1}{2}mv^2$

If the mass and velocity remain constant, kinetic energy remains constant.

The position of the object does not directly change its kinetic energy.

Q6. Does potential energy change during horizontal and vertical motion?

Answer: Potential energy does not change during horizontal motion at the same height, but it increases when the object is raised vertically.

Explanation:
Gravitational potential energy near Earth’s surface is:

$U = mgh$

If an object moves horizontally, its height $h$ does not change. So, potential energy remains the same.

If the object is gradually raised upward, its height increases. Therefore, potential energy increases.

Q7. Find mechanical energy just before a falling ball hits the ground.

Answer: Just before the ball hits the ground, its mechanical energy is still $mgh$.

Explanation:
At the top:

Potential energy = $mgh$

Kinetic energy = $0$

Mechanical energy = $mgh$

Just before hitting the ground:

Potential energy = $0$

The lost potential energy becomes kinetic energy.

Kinetic energy = $mgh$

Mechanical energy = $0 + mgh = mgh$

So, mechanical energy remains conserved if air resistance is ignored.

Q8. Explain energy changes in a ball roller coaster exhibit.

Answer: At the highest point, potential energy is maximum and kinetic energy is minimum. At lower points, kinetic energy increases and potential energy decreases.

Explanation:
At point A, the ball is at a great height. So, it has maximum potential energy.

At point B, the ball is lower and moving faster. So, potential energy decreases and kinetic energy increases.

At later points like C, D, and E, the ball usually reaches lower heights because some mechanical energy is lost due to friction and air resistance.

This lost energy changes into heat and sound.

Q9. Why are roads on hills built with gentle slopes instead of going straight up?

Answer: Roads on hills are built with gentle slopes because an inclined path reduces the force required to move upward.

Explanation:
A straight upward road would be very steep and would need a larger force.

A winding road increases the distance travelled, but reduces the effort needed to climb.

This follows the idea of an inclined plane. The work done may remain nearly the same, but the required force becomes smaller.

Q10. Why is climbing an inclined ladder easier than climbing a vertical ladder?

Answer: An inclined ladder is easier because it reduces the force needed to climb upward.

Explanation:
A vertical ladder requires a person to move almost directly against gravity.

An inclined ladder increases the distance of motion but reduces the effort needed at each step.

This is similar to an inclined plane, where a smaller force is applied over a longer distance.

Q11. Why is it easier to open a can lid using a spoon?

Answer: A spoon works as a lever and reduces the effort needed to open the lid.

Explanation:
In a lever, a small effort applied at a longer distance from the fulcrum can produce a larger force near the load.

When the spoon is used, the edge of the can acts like a fulcrum. The hand applies effort at the other end.

This increases mechanical advantage and makes the lid easier to open.

Q12. Why do we push an object closer to the scissors’ fulcrum while cutting something hard?

Answer: We push the object closer to the fulcrum to increase the cutting force.

Explanation:
Scissors work as levers. When the object is closer to the fulcrum, the load arm becomes shorter.

For the same effort, a shorter load arm gives a larger force at the cutting point.

This is why hard materials are easier to cut near the hinge of the scissors.

Q13. Why do perpetual motion machines not work?

Answer: Perpetual motion machines do not work because real machines lose energy due to friction, air resistance, sound, and heat.

Explanation:
A machine cannot create energy. It can only transfer or transform energy.

In real machines, some mechanical energy always changes into heat or sound. Because of this energy loss, the machine slows down and eventually stops unless more energy is supplied.

So, no real machine can keep doing useful work forever without fuel or energy input.

NCERT Solutions for Class 9 Science Exploration Chapter 7 Exercise Questions

The Revise, Reflect, Refine section includes true-false questions, formula-based fill-ups, energy transformation, work-energy theorem, graph-based energy questions, and simple machine applications.

Q1. State whether True or False.

Answer:

Q2. Fill in the blanks.

Answer:

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

Q3. When a ball thrown upward reaches its highest point, which statements are correct?

Answer: The correct statements are (iii) and (iv).

Explanation:
At the highest point, the ball’s velocity becomes zero for an instant. Therefore, its kinetic energy is zero.

Its height is maximum, so its potential energy is maximum.

However, the force of gravity still acts downward. So, force is not zero.

Acceleration due to gravity is also not zero. It remains $g$ downward.

Q4. Identify the energy transformation in each situation.

Answer:

Q5. Compare potential energy gained by elevator and staircase.

Question: A student of mass $50 , kg$ is lifted to the top of a building of height $72.5 , m$. Take $g = 10 , m , s^{-2}$.

Q5(i). Find gain in potential energy when lifted straight up.

Answer: The gain in potential energy is $36250 , J$.

Explanation:

$U = mgh$

$U = 50 \times 10 \times 72.5$

$U = 36250 , J$

Q5(ii). Find gain in potential energy when the student climbs stairs.

Answer: The gain in potential energy is also $36250 , J$.

Explanation:
The final height is the same. Potential energy depends on height, not on the path taken.

$U = mgh = 36250 , J$

Q5(iii). What is the conclusion?

Answer: Potential energy depends on vertical height, not on the path taken.

Explanation:
Whether the student goes straight up or climbs stairs, the final height is the same. So, the gain in potential energy is the same.

Q6. Compare energy and power needed to lift the same mass to the 10th and 20th floors.

Answer: Lifting the mass to the 20th floor requires twice the energy but the same power if the time is doubled.

Explanation:
Assume each floor has equal height.

Height of 20th floor is twice the height of 10th floor.

Potential energy gained:

$U = mgh$

If height doubles, energy required also doubles.

For power:

$P = \frac{W}{t}$

The work doubles, but the time also doubles.

So, power remains the same.

Q7. What factors determine energy and power required to raise a flag using a pulley?

Answer: Energy depends on the mass of the flag and the height raised. Power depends on how fast the flag is raised.

Explanation:
Work done in raising the flag:

$W = mgh$

So, energy required depends on:

1. Mass of the flag
2. Height of the flagpole
3. Acceleration due to gravity

Raising the flag slowly or quickly does not change the total work done if the same height is reached.

Power is:

$P = \frac{W}{t}$

If the speed is doubled, the time is halved. So, the power required becomes double.

Q8. Find the ratio of fuel used when a man rides alone and with his son.

Question: A man of mass $60 , kg$ rides a scooter of mass $100 , kg$. Next day, his son of mass $40 , kg$ joins him. The scooter reaches the same speed in the same time. Find the ratio of fuel used.

Answer: The ratio of fuel used on the two days is $4 : 5$.

Explanation:
Energy required to reach the same velocity depends on kinetic energy:

$K = \frac{1}{2}mv^2$

Day 1 total mass:

$60 + 100 = 160 , kg$

Day 2 total mass:

$60 + 100 + 40 = 200 , kg$

Since final speed is the same, fuel used is proportional to mass.

Ratio:

$160 : 200 = 4 : 5$

So, the fuel used on the first day and second day is in the ratio $4 : 5$.

Q9. Draw and explain a balanced seesaw with adult twice the child’s weight.

Answer: The child should sit twice as far from the fulcrum as the adult.

Explanation:
For a balanced seesaw:

$Effort \times effort , arm = Load \times load , arm$

If the adult weighs twice as much as the child, the adult must sit at half the child’s distance from the fulcrum.

Example:

Child at $2 , m$ from fulcrum

Adult at $1 , m$ from fulcrum

Then:

$W \times 2 = 2W \times 1$

So, the seesaw balances.

Q10. A ball of mass 2 kg is thrown up with velocity $20 , m , s^{-1}$.

Q10(i). What is the sign of work done by gravity during upward and downward motion?

Answer: During upward motion, work done by gravity is negative. During downward motion, work done by gravity is positive.

Explanation:
When the ball moves upward, gravity acts downward. Force and displacement are opposite, so work is negative.

When the ball moves downward, gravity and displacement are both downward. So, work is positive.

Q10(ii). If the ball reaches 19.4 m, find work done by air resistance.

Answer: Work done by air resistance is $-12 , J$.

Explanation:
Initial kinetic energy:

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2} \times 2 \times 20^2$

$K = 400 , J$

Potential energy at height $19.4 , m$:

$U = mgh$

$U = 2 \times 10 \times 19.4$

$U = 388 , J$

Energy lost due to air resistance:

$400 - 388 = 12 , J$

So, work done by air resistance is:

$-12 , J$

Q11. Find the block’s speed at 0 m and 4 m from the force-displacement graph.

Question: A $10.0 , kg$ block has kinetic energy $180 , J$ at $0 , m$. A variable force acts from $0 , m$ to $4 , m$. Find speed at $0 , m$ and $4 , m$.

Answer: Speed at $0 , m$ is $6 , m , s^{-1}$. Speed at $4 , m$ is $8 , m , s^{-1}$.

Explanation:
At $0 , m$:

$K = \frac{1}{2}mv^2$

$180 = \frac{1}{2} \times 10 \times v^2$

$180 = 5v^2$

$v^2 = 36$

$v = 6 , m , s^{-1}$

From the graph, force increases linearly from $0$ to $50 , N$ over $4 , m$. Work done is area under the graph:

$W = \frac{1}{2} \times 4 \times 50$

$W = 100 , J$

Final kinetic energy:

$K_f = 180 + 100 = 280 , J$

$280 = \frac{1}{2} \times 10 \times v^2$

$280 = 5v^2$

$v^2 = 56$

$v = \sqrt{56} \approx 7.48 , m , s^{-1}$

So, using the graph exactly, the speed at $4 , m$ is about $7.48 , m , s^{-1}$. If the graph is interpreted as a rectangular $50 , N$ force over $4 , m$, the work would be $200 , J$ and speed would be about $8.72 , m , s^{-1}$. The triangular graph interpretation is more likely from the figure.

The block does not have negative acceleration because the applied force is not opposite to motion. Its acceleration is positive or zero, depending on the force at each point.

Q12. How high will the ball go on the Moon?

Question: Lunar gravity is about $\frac{1}{6}$ of Earth’s gravity. An astronaut throws a ball up to $8 , m$ on Earth. How high will it go on the Moon with the same upward velocity?

Answer: The ball will rise to $48 , m$ on the Moon.

Explanation:
Maximum height for the same initial kinetic energy depends inversely on gravitational acceleration.

On Earth:

$K = mgh_E$

On Moon:

$K = mg_Mh_M$

Since $g_M = \frac{g_E}{6}$,

$mg_Eh_E = m \times \frac{g_E}{6} \times h_M$

$h_M = 6h_E$

$h_M = 6 \times 8 = 48 , m$

Q13. Answer using the speed-time graph of a car.

Q13(i). Describe how the car moves between A and B.

Answer: Between A and B, the car moves with constant speed.

Explanation:
The speed-time graph is horizontal from A to B. This means speed remains unchanged.

Q13(ii). Calculate the kinetic energy of the car at A.

Answer: The kinetic energy is $612500 , J$.

Explanation:
Mass:

$m = 1000 , kg$

Speed at A:

$v = 35 , m , s^{-1}$

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2} \times 1000 \times 35^2$

$K = 500 \times 1225$

$K = 612500 , J$

Q13(iii). State the work done by brakes in bringing the car to rest.

Answer: Work done by brakes is $-612500 , J$.

Explanation:
The brakes reduce the kinetic energy of the car to zero.

Work done by brakes:

$W = Change , in , kinetic , energy$

$W = 0 - 612500$

$W = -612500 , J$

Q13(iv). What does the kinetic energy transform into?

Answer: The kinetic energy mainly transforms into thermal energy.

Explanation:
When brakes are applied, friction between brake parts and wheels converts kinetic energy into heat. Some energy may also become sound.

Q14. Find velocity at P, Q, and R from the potential energy graph.

Question: A $0.5 , kg$ ball moves on a frictionless track. At O, velocity is $0 , m , s^{-1}$ and potential energy is $30 , J$. Find velocity at P, Q, and R.

Answer: The velocity depends on the potential energy at each point, using conservation of mechanical energy.

Explanation:
At O:

Kinetic energy = $0$

Potential energy = $30 , J$

Total mechanical energy = $30 , J$

At any point:

$K = 30 - U$

$K = \frac{1}{2}mv^2$

Since $m = 0.5 , kg$:

$\frac{1}{2} \times 0.5 \times v^2 = 30 - U$

$0.25v^2 = 30 - U$

$v = \sqrt{4(30-U)}$

Using the graph values:

Q15. Coconut falls from a 10 m tree.

Q15(i). Calculate velocity just before it hits the sand.

Answer: The velocity is approximately $14.14 , m , s^{-1}$.

Explanation:
Given:

$h = 10 , m$

$g = 10 , m , s^{-2}$

Using conservation of energy:

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 10 \times 10}$

$v = \sqrt{200}$

$v \approx 14.14 , m , s^{-1}$

Q15(ii). Calculate the depth of depression in sand.

Answer: The depth of depression is $0.05 , m$, or $5 , cm$.

Explanation:
Mass of coconut:

$m = 1.5 , kg$

Height:

$h = 10 , m$

Potential energy lost:

$U = mgh$

$U = 1.5 \times 10 \times 10$

$U = 150 , J$

This energy is used to make a depression in sand.

Work done against sand:

$W = F \times s$

$150 = 3000 \times s$

$s = \frac{150}{3000}$

$s = 0.05 , m$

So, the depression is $5 , cm$ deep.

NCERT Solutions for Class 9 Science Exploration

Topics Covered in NCERT Solutions for Class 9 Science Exploration Chapter 7

Class 9 Science Exploration Chapter 7 covers work, energy, power, and simple machines. It builds on earlier ideas of force and motion.

• Scientific meaning of work
• Work done by a constant force
• SI unit of work
• Zero work done
• Positive and negative work done
• Work-energy theorem Class 9
• Energy as capacity to do work
• Forms of energy
• Mechanical energy
• Kinetic energy Class 9
• Potential energy Class 9
• Gravitational potential energy
• Conservation of mechanical energy Class 9
• Energy changes in pendulum and roller coaster
• Power Class 9 Science
• Watt and horsepower
• Simple machines Class 9
• Pulley, inclined plane and lever Class 9
• Mechanical advantage Class 9

Important Concepts in NCERT Solutions for Class 9 Science Exploration Chapter 7

Work Energy and Simple Machines Class 9 connects force, displacement, motion, and energy transfer. These concepts help students solve numericals and explain real-life examples.

Important Formulas in NCERT Solutions for Class 9 Science Exploration Chapter 7

Class 9 Science Chapter 7 solutions use formulas for work, kinetic energy, potential energy, power, and mechanical advantage.