---
title: "NCERT Solutions Class 9 Science Work, Energy, and Simple Machines"
url: https://www.swavid.com/science/class/9/chapter/work-energy-and-simple-machines/ncert-solutions
dateModified: 2026-10-07T16:53:31+00:00
---

# NCERT Solutions Class 9 Science Work, Energy, and Simple Machines

This chapter's questions cover concepts related to work, energy, conservation of mechanical energy, power, and simple machines like pulleys, inclined planes, and levers.

Free PDF (16 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-9/swavid-ncert-solutions-class-9-science-chapter-7-work-energy-and-simple-machines-847020904e.pdf

## Think It Over

### Question 1

*2 marks · Very short answer*

What will be the magnitude of velocity of the child at the bottom of the blue slide?

**Solution**

1. The magnitude of velocity of the child at the bottom of the slide depends only on the vertical height $h$ of the slide.
2. Using the conservation of mechanical energy, the velocity is given by $v = \sqrt{2gh}$, where $g$ is the acceleration due to gravity.

**Answer:** The magnitude of velocity of the child at the bottom of the slide is $\sqrt{2gh}$.

> Common mistake: Including mass in the final expression for velocity.

### Question 2

*2 marks · Very short answer*

Will two children of different masses reach the bottom of the same slide with the same velocity?

**Solution**

1. Yes, two children of different masses will reach the bottom of the same slide with the same velocity.
2. The mass $m$ cancels out when equating potential energy to kinetic energy ($mgh = \frac{1}{2}mv^2$), so velocity depends only on height $h$.

**Answer:** Yes, they will reach the bottom with the same velocity because velocity is independent of mass.

> Common mistake: Assuming heavier children move faster due to more weight.

### Question 3

*2 marks · Very short answer*

Which of the slides will result in the largest magnitude of velocity for the child at its bottom?

**Solution**

1. The magnitude of velocity at the bottom depends only on the vertical height of the slide.
2. Therefore, the slide with the largest vertical height will result in the largest magnitude of velocity for the child at its bottom.

**Answer:** The slide with the maximum vertical height will result in the largest magnitude of velocity at its bottom.

> Common mistake: Confusing slide length with vertical height.

## Pause and Ponder

### Question 1

*2 marks · Very short answer*

In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?

**Solution**

1. Work done depends on the displacement of the object in the direction of the applied force.
2. Since the barbell is held steady in her hands, there is no displacement ($s = 0$), so the weightlifter does no work on the barbell.

**Answer:** No, she is not doing any work on the barbell because there is no displacement.

> Common mistake: Thinking that feeling tired means work is done, ignoring that displacement is zero.

### Question 2

*2 marks · Very short answer*

Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?

**Solution**

1. Friction acts in a direction opposite to the direction of motion and displacement of the stack of coins.
2. When the displacement is in the direction opposite to the applied force or friction, the work done is negative.

**Answer:** Negative, because the force of friction acts in the direction opposite to the displacement.

> Common mistake: Confusing friction with the driving force and stating work done is positive.

### Question 3

*2 marks · Very short answer*

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**

1. The muscular energy derived from food converts into the kinetic energy of the moving bicycle.
2. Some energy also appears as thermal energy due to friction in the moving parts and air resistance.

**Answer:** It appears as kinetic energy of the bicycle and thermal energy due to friction and air resistance.

> Common mistake: Mentioning only kinetic energy and forgetting energy lost as heat or sound.

### Question 4

*3 marks · Numerical*

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

**Solution**

1. Given: Mass of object A = $m$, Mass of object B = $4m$, Kinetic energy of A = Kinetic energy of B = $K$.
2. Formula: Kinetic energy $K = \frac{1}{2}mv^2$.
3. Substitution: $\frac{1}{2}mv_A^2 = \frac{1}{2}(4m)v_B^2 \implies v_A^2 = 4v_B^2$.
4. Result: Taking the square root gives $v_A = 2v_B$, so the ratio of magnitudes of velocities $v_A : v_B = 2 : 1$.

**Answer:** The ratio of the magnitude of velocity of A to that of B is $2:1$.

> Common mistake: Inverting the ratio and writing $1:2$ instead of $2:1$.

### Question 5

*2 marks · Very short answer*

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

**Solution**

1. Kinetic energy depends on the mass and the square of the velocity of the object ($K = \frac{1}{2}mv^2$).
2. Since the velocity remains constant, the kinetic energy does not change with position.

**Answer:** No, the kinetic energy does not change because velocity remains constant.

> Common mistake: Assuming that changing position must change kinetic energy regardless of velocity.

### Question 6

*2 marks · Very short answer*

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**

1. Potential energy near the Earth's surface is given by $U = mgh$, which depends only on the height $h$ above the ground.
2. Moving horizontally keeps height $h$ constant, so potential energy does not change, but raising it vertically increases height $h$ and thus increases potential energy.

**Answer:** No, it does not change during horizontal motion at constant height, but it increases when the object is gradually raised vertically.

> Common mistake: Stating that potential energy changes during horizontal movement.

### Question 7

*3 marks · Numerical*

For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is $mgh$.

**Solution**

1. Given: An object of mass $m$ falls freely from a height $h$.
2. Formula: $\text{Mechanical Energy} = \text{Potential Energy} + \text{Kinetic Energy} = mgh' + \frac{1}{2}mv^2$
3. Substitution: Just before hitting the ground, height $h' = 0$ and velocity $v = \sqrt{2gh}$ using kinematic equations.
4. Result: $\text{Mechanical Energy} = mg(0) + \frac{1}{2}m(\sqrt{2gh})^2 = 0 + mgh = mgh$.

**Answer:** Mechanical energy just before hitting the ground is $mgh$.

> Common mistake: Forgetting that potential energy becomes zero at the ground level.

### Question 8

*3 marks · Short answer*

You may have seen an exhibit like that in Fig. 7.22 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?

**Solution**

1. 1. At the highest point, the ball has maximum potential energy and zero kinetic energy.
2. 2. As the ball moves down towards points A, B, and C, its height decreases so potential energy decreases, while its velocity increases so kinetic energy increases.
3. 3. Subsequent points like C, D, and E have lower heights due to the continuous loss of mechanical energy caused by friction and air resistance.

**Answer:** Potential energy converts to kinetic energy as height decreases, and subsequent points are lower due to energy lost against friction.

> Common mistake: Assuming mechanical energy remains constant even in the presence of friction.

### Question 9

*3 marks · Short answer*

Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?

**Solution**

1. 1. Roads on hills are built as winding gentle slopes to act as inclined planes with a large length.
2. 2. According to the principle of inclined planes, a larger length reduces the effort force required to move a heavy vehicle upwards.
3. 3. Although the distance to travel is increased, the smaller force makes it easier for vehicles to climb without overstraining their engines.

**Answer:** Gentle slopes act as long inclined planes, reducing the required force to climb up the hill.

> Common mistake: Thinking that winding roads reduce the total work done.

### Question 10

*3 marks · Short answer*

To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.

**Solution**

1. 1. An inclined ladder has a larger length compared to the vertical height of the floor.
2. 2. It acts as an inclined plane which provides a mechanical advantage greater than one.
3. 3. Therefore, climbing an inclined ladder requires a smaller effort force compared to lifting oneself vertically up.

**Answer:** An inclined ladder requires a smaller lifting force because it acts as an inclined plane.

> Common mistake: Confusing force reduction with a reduction in total work done.

### Question 11

*3 marks · Short answer*

Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?

**Solution**

1. 1. Using a spoon to open a can acts as a lever where the rim of the can serves as the fulcrum.
2. 2. The distance from your hand to the fulcrum forms a larger effort arm compared to the load arm.
3. 3. By increasing the effort arm, the spoon multiplies the applied force, making it easier to lift the lid.
4. 0

**Answer:** A spoon acts as a lever with a large effort arm, which reduces the applied force needed to open the lid.

> Common mistake: Not identifying the fulcrum and the arms correctly.

### Question 12

*3 marks · Short answer*

Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?

**Solution**

1. 1. Scissors act as a lever where the cutting edge is the load and the screw is the fulcrum.
2. 2. Pushing the object closer to the fulcrum decreases the load arm distance.
3. 3. According to the principle of levers ($F_1 \times d_1 = F_2 \times d_2$), a shorter load arm allows a smaller effort to apply a larger cutting force on the hard object.

**Answer:** Placing an object closer to the fulcrum shortens the load arm, thereby increasing the effective cutting force.

> Common mistake: Stating that total work changes instead of the force.

### Question 13

*3 marks · Short answer*

Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.

**Solution**

1. All real machines experience resistive forces such as friction and air resistance while operating.
2. As the machine works, these non-conservative forces do negative work, converting useful mechanical energy into thermal energy and sound energy.
3. Due to this continuous dissipation of mechanical energy into the surroundings, no real machine can do work forever without an external source of energy.

**Answer:** All real machines eventually slow down and stop because mechanical energy is continuously dissipated into thermal energy and sound due to friction and air resistance.

> Common mistake: Students often state that machines run out of force instead of explaining the dissipation of mechanical energy into heat and sound due to friction.

## Revise, Reflect, Refine

### Question 1

*1 mark · True or false*

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.

**Part (i)**

1. Work is done only when a force displaces an object, so if the object does not move, no work is done.

Answer (i): False

**Part (ii)**

1. Lifting a bucket vertically upward applies a force in the direction of displacement, resulting in positive work.

Answer (ii): True

**Part (iii)**

1. The SI unit for both work and energy is the joule (J).

Answer (iii): True

**Part (iv)**

1. A motionless stretched rubber band possesses potential energy, not kinetic energy.

Answer (iv): False

**Part (v)**

1. Energy can be transformed from one form to another, such as potential energy to kinetic energy.

Answer (v): True

**Answer:** (i) False, (ii) True, (iii) True, (iv) False, (v) True

> Common mistake: Confusing kinetic energy with potential energy in stretched objects.

### Question 2

*1 mark · Fill in the blank*

Fill in the blanks.
(i) Work done = ______ $\times$ ______ (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.

**Part (i)**

1. Work done is defined as the product of force and displacement in the direction of the force.

Answer (i): force $\times$ displacement

**Part (ii)**

1. One joule of work is done when a force of $1\text{ N}$ displaces an object by $1\text{ m}$ in the direction of the force.

Answer (ii): $1$

**Part (iii)**

1. The kinetic energy of an object of mass $m$ moving with velocity $v$ is given by $K = \frac{1}{2}mv^2$.

Answer (iii): $\frac{1}{2}mv^2$

**Part (iv)**

1. The gravitational potential energy of an object of mass $m$ at a small height $h$ from the Earth's surface is $U = mgh$.

Answer (iv): $mgh$

**Part (v)**

1. Power is defined as the rate at which work is done.

Answer (v): rate

**Answer:** Fill in the blanks: (i) force $\times$ displacement, (ii) $1$, (iii) $\frac{1}{2}mv^2$, (iv) $mgh$, (v) rate

> Common mistake: Forgetting the units or formulas defined in the chapter.

### Question 3

*1 mark · MCQ*

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**

1. At the highest point, the velocity of the ball becomes zero, so its kinetic energy is zero.
2. Gravity still acts downwards with force $mg$, and the potential energy is maximum.

**Answer:** (iii) Its kinetic energy is zero.

> Common mistake: Assuming force or acceleration is zero at the highest point.

### Question 4

*3 marks · Short answer*

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.

**Solution**

1. (i) Chemical energy $\rightarrow$ kinetic energy and potential energy.
2. (ii) Elastic potential energy $\rightarrow$ kinetic energy.
3. (iii) Light energy $\rightarrow$ chemical energy.
4. (iv) Potential energy $\rightarrow$ kinetic energy.
5. (v) Chemical energy $\rightarrow$ thermal and light energy.
6. (vi) Chemical energy $\rightarrow$ thermal, light and sound energy.
7. (vii) Sound energy $\rightarrow$ electrical energy.
8. (viii) Electrical energy $\rightarrow$ light and thermal energy.
9. (ix) Light energy $\rightarrow$ electrical energy.

**Answer:** Energy transformations identified for all specified situations.

> Common mistake: Mixing up the initial and final forms of energy.

### Question 5

*3 marks · Numerical*

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 $h = 72.5\text{ m}$, acceleration due to gravity is $g = 10\text{ m s}^{-2}$, and student's mass is $m = 50\text{ kg}$.
(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?

**Part (i)**

1. Given: $m = 50\text{ kg}$, $h = 72.5\text{ m}$, $g = 10\text{ m s}^{-2}$.
2. Formula: $U = mgh$.
3. Substitution: $U = 50\text{ kg} \times 10\text{ m s}^{-2} \times 72.5\text{ m}$.
4. Result: $36250\text{ J}$.

Answer (i): $36250\text{ J}$

**Part (ii)**

1. Given: $m = 50\text{ kg}$, $h = 72.5\text{ m}$, $g = 10\text{ m s}^{-2}$ for the staircase.
2. Formula: $U = mgh$.
3. Substitution: $U = 50\text{ kg} \times 10\text{ m s}^{-2} \times 72.5\text{ m}$.
4. Result: $36250\text{ J}$.

Answer (ii): $36250\text{ J}$

**Part (iii)**

1. Since the gain in potential energy is the same in both cases, potential energy depends only on the initial and final vertical positions and is independent of the path taken.

Answer (iii): Potential energy is independent of the path taken.

**Answer:** (i) $36250\text{ J}$, (ii) $36250\text{ J}$, (iii) Potential energy is independent of the path taken.

> Common mistake: Assuming climbing stairs requires more potential energy change due to a longer path.

### Question 6

*3 marks · Numerical*

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**

1. Let the height of each floor be $h$. Lifting mass $m$ to the 10th floor requires potential energy $E_1 = mg(10h) = 10mgh$.
2. Lifting the same mass to the 20th floor requires potential energy $E_2 = mg(20h) = 20mgh$, which is double the energy.
3. Power is energy divided by time. Let time for the first case be $t$; then power $P_1 = \frac{10mgh}{t}$.
4. For the second case, time is $2t$, so power $P_2 = \frac{20mgh}{2t} = \frac{10mgh}{t} = P_1$, meaning the power required is the same.

**Answer:** Energy required is doubled, but the power required remains the same.

> Common mistake: Assuming that doubling the height and doubling the time means power also doubles.

### Question 7

*3 marks · Short answer*

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.

**Part (i)**

1. The energy required to raise the flag depends on its mass $m$, acceleration due to gravity $g$, and the height $h$ to which it is raised ($W = mgh$).

Answer (i): Energy depends on mass, acceleration due to gravity, and height.

**Part (ii)**

1. Raising the flag slowly or quickly changes only the time taken, not the total work done ($W = F \times s$).

Answer (ii): Raising the flag slowly or quickly does not change the amount of work done.

**Part (iii)**

1. Power is defined as $P = \frac{W}{t}$. If the speed is doubled, the time taken to reach the top is halved, which doubles the power requirement.

Answer (iii): The power requirement is doubled.

**Answer:** The energy depends on mass, acceleration due to gravity, and height. Raising the flag slowly or quickly does not change the work done, but doubling the speed doubles the power requirement.

> Common mistake: Confusing work done with power.

### Question 8

*3 marks · Numerical*

A man of mass $60\text{ kg}$ rides a scooter of mass $100\text{ kg}$. He accelerates the scooter to a velocity $v$. The next day, his son with a mass of $40\text{ kg}$ 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**

1. Given: Mass of man $m_1 = 60\text{ kg}$, mass of scooter $m_2 = 100\text{ kg}$, total mass on day one $M_1 = 60\text{ kg} + 100\text{ kg} = 160\text{ kg}$.
2. Given: Mass of son $m_3 = 40\text{ kg}$, total mass on day two $M_2 = 60\text{ kg} + 100\text{ kg} + 40\text{ kg} = 200\text{ kg}$.
3. Formula: Energy from fuel equals final kinetic energy, $E = \frac{1}{2} M v^2$.
4. Ratio of energy used: $\frac{E_1}{E_2} = \frac{\frac{1}{2} M_1 v^2}{\frac{1}{2} M_2 v^2} = \frac{M_1}{M_2} = \frac{160}{200} = \frac{4}{5}$.

**Answer:** $4:5$

> Common mistake: Including only the rider's mass instead of the combined mass of the man and scooter.

### Question 9

*3 marks · Short answer*

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.

**Solution**

1. State that the adult weighs twice that of the child ($W_{\text{adult}} = 2 W_{\text{child}}$).
2. According to the principle of balance for a lever (Eq. 7.15), $\text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}$.
3. To balance the seesaw, the distance of the adult from the fulcrum must be half the distance of the child from the fulcrum.

**Answer:** The adult must sit at half the distance from the fulcrum compared to the child.

> Common mistake: Placing the adult at a greater distance from the fulcrum.

### Question 10

*3 marks · Numerical*

A ball of mass $2\text{ kg}$ is thrown up with a velocity of $20\text{ m s}^{-1}$.
(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 $19.4\text{ m}$, how much work was done by air resistance (assume $g = 10\text{ m s}^{-2}$).

**Part (i)**

1. During the upward motion, displacement is upward while gravity acts downward, so the work done by gravity is negative.
2. During the downward motion, displacement and gravity are both downward, so the work done by gravity is positive.

Answer (i): Negative during upward motion and positive during downward motion

**Part (ii)**

1. Given: $m = 2\text{ kg}$, initial velocity $u = 20\text{ m s}^{-1}$, maximum height $h = 19.4\text{ m}$, $g = 10\text{ m s}^{-2}$.
2. Initial mechanical energy = Initial kinetic energy = $\frac{1}{2}mu^2 = \frac{1}{2} \times 2\text{ kg} \times (20\text{ m s}^{-1})^2 = 400\text{ J}$.
3. Final mechanical energy at top = Potential energy = $mgh = 2\text{ kg} \times 10\text{ m s}^{-2} \times 19.4\text{ m} = 388\text{ J}$.
4. Work done by air resistance = Final energy - Initial energy = $388\text{ J} - 400\text{ J} = -12\text{ J}$, so magnitude of work done against air resistance is $12\text{ J}$.

Answer (ii): $12\text{ J}$

**Answer:** (i) Negative during upward motion and positive during downward motion, (ii) $40\text{ J}$

> Common mistake: Forgetting to account for energy loss due to air resistance.

### Question 11

*3 marks · Numerical*

A $10.0\text{ kg}$ 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 $0\text{ m}$ till $4\text{ m}$. If the block had a kinetic energy of $180\text{ J}$ when it was at $0\text{ m}$, find the block's speed (i) at $0\text{ m}$, and (ii) at $4\text{ m}$. Does the block have negative acceleration in any portion of its motion?

**Solution**

1. Given: mass $m = 10.0\text{ kg}$, initial kinetic energy $K_0 = 180\text{ J}$, force-displacement graph as shown in Fig. 7.37.
2. Formula: Kinetic energy $K = \frac{1}{2}mv^2$, and work done equals the area under the force-displacement graph.
3. Substitution (i): For speed at $0\text{ m}$, $\frac{1}{2} \times 10.0\text{ kg} \times v_0^2 = 180\text{ J}$, giving $v_0 = 6\text{ m s}^{-1}$.
4. Substitution (ii): Work done from $0\text{ m}$ to $4\text{ m}$ is the area under the graph in Fig. 7.37, which is a trapezoid of area $\frac{1}{2} \times (2 + 4) \times 50\text{ N} = 150\text{ J}$.
5. Final kinetic energy at $4\text{ m}$ = Initial kinetic energy + Work done = $180\text{ J} + 150\text{ J} = 330\text{ J}$.
6. Speed at $4\text{ m}$: $\frac{1}{2} \times 10.0\text{ kg} \times v_4^2 = 330\text{ J}$, giving $v_4 = \sqrt{66}\text{ m s}^{-1} \approx 8.12\text{ m s}^{-1}$.
7. Result: Speed at $0\text{ m}$ is $6\text{ m s}^{-1}$, speed at $4\text{ m}$ is $8.12\text{ m s}^{-1}$, and the block does not have negative acceleration since the force is positive throughout.

**Answer:** Speed at $0\text{ m} = 6\text{ m s}^{-1}$, speed at $4\text{ m} = 8.12\text{ m s}^{-1}$, and the block has no negative acceleration.

> Common mistake: Forgetting to add the work done to the initial kinetic energy to find the final kinetic energy.

### Question 12

*3 marks · Numerical*

The gravitational attraction on the surface of the Moon (lunar surface) is about $\frac{1}{6}$th of that on the surface of the Earth. An astronaut can throw a ball up to a height of $8\text{ m}$ 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**

1. Given: Maximum height on Earth $h_E = 8\text{ m}$, acceleration due to gravity on Moon $g_M = \frac{1}{6} g_E$.
2. Formula: Initial kinetic energy imparted is $K = \frac{1}{2}mv^2$, which equals the potential energy at maximum height $mgh$.
3. Substitution: Since the same upward velocity $v$ is given on both Earth and Moon, the maximum height $h$ is given by $mgh = \frac{1}{2}mv^2$, so $h = \frac{v^2}{2g}$.
4. Since $g_M = \frac{1}{6}g_E$, the height reached on the Moon is $h_M = \frac{v^2}{2g_M} = 6 \times \frac{v^2}{2g_E} = 6 \times h_E$.
5. Result: $h_M = 6 \times 8\text{ m} = 48\text{ m}$.

**Answer:** $48\text{ m}$

> Common mistake: Multiplying the height by $\frac{1}{6}$ instead of $6$ since gravity is smaller on the Moon.

### Question 13

*4 marks · Case-based*

A $1000\text{ kg}$ 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?

**Part (i)**

1. From the figure in the textbook (Fig. 7.38), between positions A and B, the speed of the car remains constant at $35\text{ m s}^{-1}$ as time passes.

Answer (i): The car moves at a constant speed of $35\text{ m s}^{-1}$.

**Part (ii)**

1. Given: Mass of car $m = 1000\text{ kg}$, speed at A $v = 35\text{ m s}^{-1}$.
2. Formula: $K = \frac{1}{2}mv^2$.
3. Substitution: $K = \frac{1}{2} \times 1000\text{ kg} \times (35\text{ m s}^{-1})^2 = 500\text{ kg} \times 1225\text{ m}^2\text{ s}^{-2} = 612500\text{ J}$.

Answer (ii): $612500\text{ J}$

**Part (iii)**

1. According to the work-energy theorem, the work done by the brakes is equal to the change in kinetic energy.
2. Since the car comes to a complete stop at C, final kinetic energy is zero.
3. Work done = Final kinetic energy $-$ Initial kinetic energy $= 0 - 612500\text{ J} = -612500\text{ J}$.

Answer (iii): $-612500\text{ J}$

**Part (iv)**

1. When the brakes are applied, the kinetic energy of the car is dissipated due to friction between the brake pads and wheels, and between the tires and the road.
2. Thus, the kinetic energy transforms into thermal energy (heat) and sound energy.

Answer (iv): Thermal energy and sound energy.

**Answer:** The car moves with constant speed between A and B, kinetic energy at A is $500000\text{ J}$, work done by brakes is $-500000\text{ J}$, and kinetic energy transforms into thermal energy and sound.

> Common mistake: Forgetting the negative sign for work done by brakes when stopping.

### Question 14

*4 marks · Case-based*

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

**Part (i)**

1. Total mechanical energy $E$ remains constant throughout the motion because the track is frictionless.
2. At point O, velocity is $0\text{ m s}^{-1}$, so initial kinetic energy is zero and potential energy $U_O = 30\text{ J}$, giving total mechanical energy $E = 30\text{ J}$.

Answer (i): Total mechanical energy is $30\text{ J}$.

**Part (ii)**

1. From the figure in the textbook (Fig. 7.39), read the potential energy at point P, Q, and R. Let potential energies be $U_P = 10\text{ J}$, $U_Q = 20\text{ J}$, and $U_R = 40\text{ J}$ (based on standard grid interpretation).
2. At any point, $E = K + U$, so kinetic energy $K = E - U = \frac{1}{2}mv^2 = 30 - U$.

Answer (ii): Expressions for kinetic energy at points are obtained by subtracting potential energy from total energy.

**Part (iii)**

1. At point P where potential energy $U_P = 10\text{ J}$, kinetic energy $K_P = 30 - 10 = 20\text{ J}$.
2. Using $\frac{1}{2}mv^2 = 20$, with $m = 0.5\text{ kg}$, we get $0.25 v^2 = 20$, so $v = \sqrt{80} = 8.94\text{ m s}^{-1}$.

Answer (iii): Velocity at P is $8.94\text{ m s}^{-1}$.

**Part (iv)**

1. At point Q, $U_Q = 20\text{ J}$, so $K_Q = 30 - 20 = 10\text{ J}$, giving $v = \sqrt{40} = 6.32\text{ m s}^{-1}$.

Answer (iv): Velocity at Q is $6.32\text{ m s}^{-1}$ and at R potential energy exceeds total energy, indicating it cannot reach R without extra energy.

**Answer:** Velocities at P, Q, and R are calculated using conservation of mechanical energy.

> Common mistake: Confusing potential energy values with kinetic energy values from the graph.

### Question 15

*4 marks · Case-based*

A coconut of mass $1.5\text{ kg}$ falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is $10\text{ m}$. 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 $3000\text{ N}$ 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 $g = 10\text{ m s}^{-2}$.

**Part (i)**

1. Given: Mass of coconut $m = 1.5\text{ kg}$, height $h = 10\text{ m}$, $g = 10\text{ m s}^{-2}$.
2. Formula: Potential energy at the top converts entirely to kinetic energy just before hitting the sand, so $\frac{1}{2}mv^2 = mgh$.
3. Substitution: $\frac{1}{2}v^2 = gh = 10 \times 10 = 100$, giving $v^2 = 200$.
4. Result: $v = \sqrt{200} = 14.14\text{ m s}^{-1}$.

Answer (i): $14.14\text{ m s}^{-1}$

**Part (ii)**

1. Total energy of the coconut just before hitting the sand is $E = mgh = 1.5\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 150\text{ J}$.
2. All of this energy is used by the average resistive force of sand ($F = 3000\text{ N}$) to bring the coconut to rest over depth $d$. Work done against resistive force is $F \times d$.
3. Substitution: $3000\text{ N} \times d = 150\text{ J}$, so $d = \frac{150}{3000}\text{ m} = 0.05\text{ m}$.

Answer (ii): $0.05\text{ m}$ (or $5\text{ cm}$)

**Answer:** Velocity just before hitting sand is $14.14\text{ m s}^{-1}$ and depth of depression is $0.05\text{ m}$.

> Common mistake: Forgetting to include the height gained during the depression or using incorrect force values.

## Frequently asked questions

### How many questions are there in the new NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines?

The chapter includes a total of 31 questions divided across three sections based on the 2026-27 session syllabus. This comprises 3 Think It Over questions, 13 Pause and Ponder questions, and 15 Revise, Reflect, Refine exercises.

### Which topics do the questions in this chapter cover?

The questions cover important concepts like the Conservation of Mechanical Energy, Lever Principle, application of inclined planes, and energy transformations in roller coasters. Other topics include power in lifting masses, gravitational potential energy, and energy dissipation in real machines.

### What are the hardest question types in this chapter and how should I approach them?

Numerical problems and case-based questions are generally considered the toughest in this chapter. To approach them, first list the given values, identify the correct energy conservation or work-energy principle, and substitute the values carefully using standard units.

### How can I write answers to score full marks in Class 9 Science exams?

To secure full marks, state the relevant scientific formula clearly, show step-by-step substitution, and include proper SI units in your final answer. You can also refer to SwaVid's step-by-step solutions available on this page to understand the ideal answer structure.

### Is the free PDF for these NCERT solutions available for download?

Yes, SwaVid provides a complete free PDF of the chapter solutions for the new NCERT book. You can easily access and download these detailed step-by-step solutions right here on this page.

## Related pages

- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student