Kerala Syllabus Class 9 Physics Chapter 6 Work and Energy Notes Solutions

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 6 Work and Energy Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 6 Notes Solutions Work and Energy

SCERT Class 9 Physics Chapter 6 Notes Solutions Kerala Syllabus Work and Energy Questions and Answers

Class 9 Physics Chapter 6 Let Us Assess Answers Work and Energy

Question 1.
Which one of the following quantities has the same unit as that of work?
a) Power
b) Energy
c) Force
d) Displacement
Answer:
Energy

Question 2.
Which of the following pairs are scalar quantities?
a) Time and energy
b) Force and power
c) Speed and acceleration
d) Displacement and velocity
Answer:
Time and energy

Question 3.
Which of the following is the unit of power in terms of the fundamental units metre, kilogram and second?
a) Kgm²/s³
b) Kgm/s
c) Kgm/s²
d) Kgm²/s
Answer:
Kgm²/s³

Question 4.
An object of mass 2.4 kg is kept on a level surface. On applying a force of 50 N, the object moves 8 m in the direction of the force. Calculate the quantity of work done.
a) 40 J
b) 400 J
c) 50 J
d) 17.6 J
Answer:
Mass, m = 2.4 kg
Force, F = 50 N
Displacement, s = 8 m
Work done, w = F × s = 50 × 8 = 400 J

Question 5.
The figure depicts a body of mass 20 kg lifted to a height 5m using a pulley. If the work done is 1020 J, answer the following questions. (Acceleration due to gravity = 10m/s²)

a) What is the work done here if the force of friction of the pulley is not considered?
b) Calculate the work done against the friction of the pulley.
Answer:
a) mass, m = 20 kg
Height, h = 5 m
Work done = 1020J
Work done = F × s
The force required to lift the body is equal to the weight of the body.
Weight = m = 20 × 10 = 200 N
Work done to lift the body to a height of 5 meters
W = F × s = 200 × 5 = 1000 J

b) Work done against friction = Total work done – work done to lift without friction = 1020 – 100 = 20 J

Question 6.
An electric motor pumps 186.5 kg of water into a tank at a height of 8m in 10 s (g = 10 m/s²)
a) Calculate the power of the motor.
b) Express this in horse power.
c) How much time will it take to fill the same quantity of water if the power of the motor is halved?
Answer:
a) m = 186.5 kg, h = 8 m, t = 10 s
Power = \(\frac{W}{t}\)
Work = F × s
Force is equal to the weight of the body
Weight= m × g
= 186.5 × 10 = 1865
Work = 1865 × 8 = 14920J
P = \(\frac{14920}{10}\) = 1492 W

b) 1 horse power = 746 W
To convert 1492W to horsepower
Horsepower = \(\frac{1492 W}{746 W/hp }\) = 2.00hp

c) Total power = 1492 W
New power = \(\frac{1492}{2}\) = 746 W
We know, Power = \(\frac{work}{time}\)
Time = \(\frac{work}{power}\) = \(\frac{14920}{746}\) = 20 s
Thus, it will take approximately 20 seconds to fill the same quantity of water if the power of the motor is halved.

Question 7.
The mass of a body is 4 kg. When a continuous force of 3 N is applied to this object towards the east, there is a displacement of 6 m in the direction of the force. Then, the same force (3 N) is continuously applied towards the south, and the displacement is 5 m in the direction of this force.
a) Calculate the work done when the object is moved towards the south.
b) Calculate the total work done on the object.
Answer:
a) Force = 3 N
Work = F × s
= 3 × 5 = 15 J
b) Work done when moving east
Work = F × s
= 3 × 6 = 18 J
Total work done = 15 + 18 = 33 J

Question 8.
If an object is thrown vertically upwards, its potential energy and kinetic energy will change. Which of the following is the correct graph related to kinetic energy? Justify your answer.

Answer:
(a)
As the object moves upwards, gravitational force does work against the upward motion, reducing its speed. As the speed decreases. Its kinetic energy gradually decreases when it reaches the maximum height. At maximum height, the velocity is equal to zero.

Question 9.
The Vikram lander module of Chandrayan III has a mass of 1752 kg.
a) What will be its potential energy when it reaches a height of 100 m from the lunar surface? (Acceleration due to gravity on the moon is 1.6 m/s²)
b) If the same Vikram module is located at a height of 100 m from the Earth’s surface, then what will be its potential energy? (g = 10 m/s²)
Answer:
a) m = 1752 kg
g = 1.6 m/s²
h = 100 m
P.E = m × g × h = 1752 × 1.6 × 100 = 280320 J

b) g = 10 m/s²
P.E = m × g × h = 1752 × 10 × 100 = 1725000 J

Question 10.
A person of mass 80 kg is standing on a suspended platform of mass 170 kg to paint the outer wall of a flat. The quantity of work done by the motor to raise him and the platform to this height is 150 kJ. How high is the platform? (g = 10 m/s²)
Answer:
Mass of person = 80 kg
Mass of suspended platform = 170kg
Work = 150 kJ = 150000J
Work = Force × displacement (Height)
In this case, the total force is the weight of both the person and the platform
Total mass = mass of person + mass of platform = 80 + 170 = 250 kg
Weight = Total mass × g = 250 × 10 = 2500 N
Height = \(\frac{work}{Force}\) = \(\frac{150000}{2500}\) = 60m

Question 11.
The figure shows two cars of the same mass and a lorry of double the mass moving along a straight road. Observe the figure and answer the following questions.

a) Will the kinetic energy of both the cars be the same? Justify your answer.
b) Is the kinetic energy of the lorry and the car in front the same? Why?
c) Calculate the kinetic energy of the car at the back if its mass is 1800 kg.
Answer:
a) No.
We know K.E = \(\frac{1}{2}\) × mv²
Here, both cars have the same mass, but they are moving with different velocities. So, they will have different kinetic energy

b) No, because the lorry and car have different masses.

c) K.E = \(\frac{1}{2}\) × mv² = \(\frac{1}{2}\) × 1800 × 20² = 360000 J.

Question 12.
Analyse the figure and answer the following questions.
A. indicate the topmost position of a building. The three stages of a ball, which is at rest falling freely from A, are depicted.

a) What is the kinetic energy at A?
b) What is the potential energy at A?
c) What is the total energy at A?
d) What is the kinetic energy at C?
e) What is the potential energy at C?
f) What is the total energy at C?
g) What change in energy has taken place as the ball falls? Describe.
Answer:
a) K.E = 0J
b) P.E = 20 J
c) Total energy at A = 20 J
d) Kinetic energy at C = 20 J
e) Potential energy at C = 0 J
f) Total energy at C = 20 J
g) Potential energy of the ball is converted into kinetic energy. According to the law of conservation of energy, the total energy remains constant.

Question 13.
Using a conveyor belt that transports construction material to the top of a building, 5 sacks of cement of mass 50 kg each are brought to a height of 8 m in 16 s. Calculate the power of the motor of the conveyor belt (g = 10 m/s²).
Answer:
mass of one cement sack = 50 kg
Mass of 5 sacks = 5 × 50 = 250 kg
We know, Power = \(\frac{Work}{Time}\)
Work = F × S
Force applied here is equal to the weight of the body
Weight W = m × g = 250 × 10 = 2500N
Then, W = 2500 × 8 = 20000 J
Therefore, P = \(\frac{20000}{16}\) = 1250 W

Question 14.
Babu and Raju have brought the materials for house construction to the top of a 15 m tall building. The details are given below.

—–	Weight	Time taken to lift the load to the height
Babu	600 N	300 s
Raju	400 N	200 s

a) Find out who did more work.
b) If it is said that Babu has more power, will you agree with it? Explain.
Answer:
a) Height of the building = 15 m

Babu
Weight = 600 N, Time = 300s
Work = F × s = 600 × 15 = 9000J

Raju
Weight = 400 N, Time=200s
Work F × s = 400 × 15 = 6000J
Babu did more work.

b) No.
Babu
Power = \(\frac{Work}{Time}\)
P = \(\frac{9000}{300}\) = 30 W
Raju
Power = \(\frac{Work}{Time}\)
P = \(\frac{6000}{200}\) = 30 W
Both have the same power.

Question 15.
Observe the graph between the force and displacement of a body.

a) From the graph, find the magnitude of the force exerted on the object.
b) From the graph, find the maximum displacement cause by the applied force.
c) What is the method to find the quantity of work using graph? Find the area under the graph and check whether it is equal to the work done.
Answer:
a) 6 N
b) 0.4 m
c) Area under Force – displacement graph gives work.
Area of rectangle = length breadth = 0.4 × 6 = 2.4 J
Work done = F × s = 0.4 × 6 = 2.4 J

Class 9 Physics Chapter 6 Extended Activities Answers Work and Energy

Question 1.
The development of science and technology has influenced the records in the field of sports. There has been an enormous change in the development and design of materials of the pole used in pole vaults, sports shoes, sportswear, etc. This has greatly influenced the results of competitions. Collect more information related to this and make a note of it.
(For example, compare the change in materials used in making the pole and the change in new heights cleared by the pole vaulters.)
Answer:
(Hints)
The development of science and technology has great importance in the world of sports. There are many changes happened in pole vaulting due to science and technology.

Bamboo

• In the early days, poles were made from materials such as bamboo or wood.
• The poles were relatively flexible and not so strong.
• This restricted the height athletes could achieve.

Fiber glass

• Later, fibreglass poles began to be used. It is very lightweight and flexible and allows vaulters to produce more energy and achieve greater heights.

Carbon fiber poles

• The latest evolution is the use of carbon fibre poles, which are even lighter and more durable than fibreglass.
• This helps athletes to clear greater heights.

Advances in Sport shoes
Early athletic shoes were made with leather and simple rubber soles, which provided less support. Today’s pole vault shoes are made with materials like lightweight synthetic fabrics and specialised rubber compounds that improve grip and energy return.

Question 2.
Analyse the chart on energy consumption across the world. If development is to continue at the same pace, energy consumption has to increase.

Prepare a seminar paper on the steps to be adopted to meet the increased demand for energy.
Answer:
(Hints)
Use renewable forms of energy
Increase the use of electric vehicles
Provide awareness among the public

Question 3.
Check the power of electric appliances in your house. Prepare short notes on steps to be taken to make the energy consumption scientific. Discuss in class and prepare a notice listing scientific methods to reduce electricity consumption.
Answer:
(Hints)
Check Appliance power: Note down the power rating of all household appliances to understand their energy usage.
Use energy-efficient appliances: Use appliances with high energy ratings to reduce consumption. Turn off when not in use: Always switch off lights and appliances when they are not needed. Switch to LED lighting: Replace traditional bulbs with energy-efficient bulbs to save energy and costs.

Question 4.
When modern cars are built, the front and rear bumpers are designed to break quickly. This is essential for the safety of passengers. Using strong crash guards at the front is a threat to the lives of the passengers.
Prepare a note explaining this based on the work-energy concepts we have learned.
Answer:
(Hints)
Modern car bumpers are designed to break during a crash to keep passengers safe. When a car hits something, it creates a lot of energy. Bumpers that break can absorb some of this energy, which helps less impact on passengers.

Work and Energy Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Analyse the situations and complete the table.

Situations where force is applied	Objects on which force is applied	Is there displacement for the object? (Yes/ No)	Is there displacement for the object in the direction of applied force? (Yes/No)
Loading goods on to a lorry	Load	Yes	Yes
Trying to push a huge rock	Rock	No
A trolley being pushed in a super market
A man standing with a load on his head

Answer:

Situations where force is applied	Objects on which force is applied	Is there displacement for the object? (Yes/ No)	Is there displacement for the object in the direction of applied force? (Yes/No)
Loading goods on to a lorry	Load	Yes	Yes
Trying to push a huge rock	Rock	No	No
A trolley being pushed in a super market	Trolley	Yes	Yes
A man standing with a load on his head	Load	No	No

From this table, it is clear that there is no displacement in the direction of the applied force in all these situations.

Question 2.
In table 1, identify the situations where work is done.
Answer:

• Loading goods onto a lorry
• A trolley being pushed in a supermarket

Question 3.
Identify the situations where work is not done.
Answer:

• Trying to push a huge rock
• A man standing with a load on his head

Question 4.
Isn’t it because the frictional force on the wheelchair increases as the patient’s weight increases?
Answer:
Yes.
Patients in wheelchairs are to be shifted to a distance of 10 m. A force of 40 N is to be applied to the wheelchair in situation 1. The force needed is 70 N in situation 2. In both cases, the wheelchair was taken at the same speed along the same path. The displacement of the wheelchairs is also the same. When we move objects across a floor on a horizontal plane, we are applying a force against the force of friction. Hence, we are doing work against the force of friction.

Question 5.
In which situations was more work done? (Situation 1/ Situation 2)
Answer:
Situation 2.

Question 6.
Here, which factor influences the quantity of work done? (force/ displacement)
Answer:
Force.
If the patient in the second situation has to be shifted through a distance of 20 m under the same conditions, the quantity of work done will increase.

Question 7.
Here, which factor influences the quantity of work done?
force/displacement
Answer:
Displacement.

Question 8.
From the situations discussed above, write down in your science diary, the factors influencing the work done.
Answer:
Force
Displacement
When a force F newton is continuously applied on a body and if the displacement in the direction of the applied force is s metre, then the work done by the force on the body will be,

Work done = force × displacement
W = Fs
The unit of work = unit of force × unit of displacement = Nm
The unit of work is newton metre. It is known as joule in SI unit, named in honour of the scientist James Prescott Joule.

Question 9.
If a force of 1 N moves an object through a distance of 1 m in the direction of force along a horizontal surface, calculate the work done by the force.
Answer:
W = F × S
Force F = 1 N
Displacement s = 1 m
Work done W = F × S = 1 × 1 = 1 Nm

Question 10.
Shouldn’t force be applied against gravity while lifting objects?
Answer:
Yes.

Question 11.
Calculate the quantity of work required to lift an object of mass 100 g to a height 1m (g = 10 m/s²).
Answer:
m = 100 g = \(\frac{100}{1000}\) kg = 0.1 kg
g = 10 m/s²
h = 1m
Work (W) = mgh = 0.1 × 10 × 1 = 1J

Question 12.
A barrel of mass 50 kg is to be loaded onto a lorry. The platform of the lorry is at a height of 2 m from the ground level (g = 10 m/s²).
a) Calculate the work done against the force of gravity when the barrel is loaded onto the lorry.
b) What is the work done against the force of gravity if it is rolled to a height of 2 m using an inclined plane of length 5 m?
c) Does the work done against gravity depend on its path of motion? yes/no
Answer:
a) m = 50 kg
g = 10 m/s²
h = 2m
Work (W) = mgh = 50 × 10 × 2 = 1000 J
b) Work done against the force of gravity
Work (W) = mgh = 50 × 10 × 2 = 1000 J
c) Work done against gravity does not depend on its path of motion.

Question 13.
Observe it and complete the table. Here, the forces acting on the object are a force of gravitation – A, the force of friction B, the force applied by the table on the box – C and the force applied by the child on the box – D.

Force experienced on the table	Direction of force
Force applied by the child	P → D
Force of gravitation
Force of friction.
Force applied by the table on the box

Answer:

Force experienced on the table	Direction of force
Force applied by the child	P → D
Force of gravitation	A → P
Force of friction.	P → B
Force applied by the table on the box	C → P

Question 14.
In which direction will the box move? (P → D, P → A, P → C, P → B)
Answer:
P→ D

Question 15.
In which direction of the force did the box get displaced?
Answer:
The displacement of the box is in the direction of the force applied by the child on the object.

Question 16.
Against which force did the box get displaced?
Answer:
The box is displaced in the opposite direction of the frictional force.

Question 17.
Why did the force of gravity and the force applied by the table cause no displacement to the box?
Answer:
The box is displaced in the direction of the force applied by the child. So, we can consider that this force has done a work on the box. Such a work is positive work. The displacement of the box is in the opposite direction to the force of friction. Here work done by the frictional force on the box is considered as negative work.

Question 18.
Write down whether work is positive or negative in each of the following situations,
a) The work done by the force of gravity on a stone when it is thrown up.
b) The work done on the stone by the applied force when we throw a stone upwards.
c) The work done by the force applied by a crane on the object while lifting it.
d) The work done on an object by a crane when the object is lifted up.
e) The work done by the force of gravity when a coconut falls down.
Answer:
a) Negative work
b) Positive work
c) Positive work
d) Positive work
e) Positive work

Question 19.
In a supermarket, a shopper pushes a trolley to a distance of 16 m by applying a force of 95 N continuously. Calculate the work done.
Answer:
F = 95 N
s = 16 m
= F × s = 95 N × 16 m = 1520 J

Question 20.
Don’t we get tired while doing various activities? Isn’t it because we utilise energy to do work?
Answer:
Yes.

Question 21.
Write down examples of different forms of energy you are familiar with.
Answer:
Electrical energy, Mechanical energy, Heat energy, Solar energy, Chemical energy, Sound energy

Question 22.
In which unit is the quantity of energy expressed?
Answer:
It is energy that is transformed into work when we do different activities. So, the unit of work is the unit of energy.
The SI unit of energy is joule (J). For indicating the quantity of energy in food, we also use the unit kilocalorie.

Question 23.
What is the relation between joule and calorie?
Answer:
1 cal (calorie) = 4.2 J
1 kcal = 4200 J
(1000 calorie = 1 kcal)

Question 24.
The quantity of energy contained in a ripe mango is approximately 840 kJ. Express this energy in calorie.
Answer:
Energy = 840 kJ = 840 × 1000 J = (840 × 1000 J)/ 4.2
= 200000 cal = 200 kcal

Question 25.
What is peculiar about their positions?
Answer:
They are in a position of strain.

Question 26.
What would happen if the arrow from the arched bow and the car at a height are released?
Answer:
They will be in motion.

Question 27.
Where does the energy for the arrow to move come from?
Answer:
The arrow gets its energy due to the unique configuration created by the tension of the string and the bend of the bow.

Question 28.
How does the toy car get energy to move?
Answer:
A toy car gets the energy it needs to move because of its position.

Question 29.
Didn’t these objects get energy due to their position or configuration?
Answer:
Yes.

Question 30.
Classify the following situations suitably on the basis of the potential energy acquired.

Answer:

Potential energy due to position	Potential energy due to configuration
Energy of coconut on the coconut tree. Energy of the water inside the tank.	Energy in a compressed spring. Energy in the bent pole during the pole vault jump.

The work done against the force of gravity is the cause of potential energy due to its position. The potential energy due to its position is equal to the quantity of work done on the object against the force of gravity.
Potential energy due to position, E_p = mgh
Here m is the mass and h is the height from the ground.

Question 31.
An object of mass 50 kg is lifted to a height of 8 m from the ground.
a) What is its potential energy when it is on the ground?
b) Calculate the potential energy when the body is at a height of 8m.
Answer:
a) Potential energy is always calculated relative to a reference level. By considering the potential energy of a body at the reference level as zero, its potential energy at each position is calculated. The ground level is taken as the reference while measuring height. Therefore, the potential energy at ground level is usually assumed as zero.

b) The mass of object, m = 50 kg
Acceleration due to gravity, g = 9.8 m/s²
The change of position of the body (displacement or height) against the force of gravity, h = 8 m.
Potential energy, E_p = mgh = 50 × 9.8 × 8 = 3920 J

Question 32.
Is it not because of their motion that wind and water are able to do work?
Answer:
Wind and water are able to do work because of their motion.

Question 33.
Will the rectangular block move farther? What could be the reason?
Answer:
Yes. It is because of the increase in mass that the rectangular box moved farther. Kinetic energy increases with an increase in mass.
Is there any other factor that affects kinetic energy?
Let’s repeat the experiment by increasing the speed of the tin.

Question 34.
Repeat the experiment by rolling down the same tin filled with soil from a greater height (by increasing the height of the inclined plane). What do you observe?
Answer:
Here, the rectangular block moved a longer distance due to the increase in velocity of the soil-filled tin.
་
Question 35.
Now, write down the two factors that influence kinetic energy.
Answer:
Mass of the object (m)
Velocity of object

Question 36.
Can Kinetic energy be calculated?
Answer:
Work, W = Fs
According to Newton’s second law of motion, F = ma. Hence work, W = Fs = (ma)s
From the equations of motion, we have v² = u² + 2 as
If initial velocity u = 0 then
v² = 2as
as = \(\frac{v^2}{2}\)
W = mas = \(\frac{\mathrm{mv}^2}{2}\) = \(\frac{1}{2}\) mv²
Kinetic energy of the object, E₁ = mv²
If m is the mass in kg and v is the velocity in m/s then E_k represents the kinetic energy in joule (J).

Question 37.
Complete the table associated with the kinetic energy of objects.

Mass of the object (m) kg	Velocity of the object (v) m/s	Kinetic energy (J) EK = 1⁄2 mv2
10	2	20
5	2
10	4
10	1

Answer:

Mass of the object (m) kg	Velocity of the object (v) m/s	Kinetic energy (J) EK = 1⁄2 mv2
10	2	20
5	2	10
10	4	80
10	1	5

Question 38.
Based on the table, write down the changes in kinetic energy in the following cases.
a) When the mass is doubled
b) When the velocity is tripled
c) When the velocity is halved
Answer:
a) When mass is doubled, kinetic energy also doubles.
b) When the velocity is tripled, kinetic energy increases by 9 times.
c) When the velocity is halved, the kinetic energy is reduced to 1/4.

Question 39.
A motorcyclist is travelling in a straight line at a speed of 10 m/s. The combined mass of the motorcycle and the passenger is 300 kg.
Answer:
a) What is the kinetic energy of the motorcycle and the passenger?
b) What is the kinetic energy if the speed is doubled?
a) Kinetic Energy = E_K = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) × 300 × 10² = 15000 J

b) Kinetic Energy = Ek = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) × 300 × 20² = 60000 J
Here kinetic energy increases by 4 times.

Question 40.
An athlete throws a shot of mass 6 kg with a speed of 12 m/s. What will be the kinetic energy of the shot?
Answer:
Kinetic Energy = E_K = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) × 6 × 12² = 432 J

Question 41.
A ball of mass 0.5 kg is at a certain height above the ground.
a) If the potential energy of the ball is 98 J, how high will the ball be? (g = 9.8 m/s²)
b) What will be the kinetic energy of the ball just before hitting the ground, if it is allowed to fall freely?
Answer:
a) m = 0.5 kg
g = 9.8 m/s²
Ep = 98 J
Ep = mgh
98 = 0.5 × 9.8 × h
h = \(\frac{98}{0.5 \times 9.8}\) = 20 m

b) v² = u² + 2as
= 0 + 2 × 9.8 × 20 = 392 J
Kinetic Energy = E_K = \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\) × 0.5 × 392 = 98 J
Here the potential energy of the ball has become kinetic energy.

Question 42.
Don’t we use a variety of machines to make it easier? Write down the names of some machines and energy conversion in them.
Answer:

• An electric motor converts electric energy into mechanical energy.
• Petrol car converts chemical energy into kinetic energy.
• Generator converts mechanical energy into electrical energy
• Cars that use electricity convert electrical energy into mechanical energy.
  

Question 43.
How does the balloon get the required energy/force to move the toy car?
Answer:
It is obtained from the force of the outflow of air from the balloon.

Question 44.
What change happens to the rubber of the balloon when it is inflated?
Answer:
The rubber in the balloon stretches and the balloon expands. The configuration of the balloon changes.

Question 45.
Which type of energy is possessed by an inflated balloon?
Answer:
Potential energy.

Question 46.
What change in energy occurs when the car moves?
Answer:
Here the potential energy of the balloon is converted into kinetic energy of the toy car.

Question 47.
Make a car like this. What are the essential features required to make your toy car better? Compare the merits of your car with those of others.
Answer:
(Hints)

• Lightweight Materials: Use light materials (plastic, paper) to reduce energy use and increase speed.
• Wheels & Axles: Ensure smooth, low-friction wheels (bottle caps, CDs) for better movement.
• Balloon Size & Placement: Use a larger balloon and direct air backward for maximum thrust. Streamline the car body to reduce air resistance.
• Wheel Alignment: Align wheels properly for straight movement.
• Stable Base: A wider base prevents toppling.

Concepts:

• Work: The balloon’s air pushes the car forward.
• Kinetic Energy: Released air turns potential energy into motion.
• Power: Efficient energy use leads to faster motion.

Comparing Cars with these:

• Speed: Better alignment, lightweight materials, and larger balloons increase speed.
• Distance: Less friction and streamlining lead to longer travel.
• Efficiency: Efficient energy use means less air needed.
• Power Output: Faster acceleration means higher power output.

Question 48.
Have you noticed the piling process involved in the construction of buildings and bridges for basement reinforcement?
Answer:
Yes.
A pit is formed as a result of the powerful hammer force from the machine hitting the pile.
The potential energy of the hammer can be calculated when the hammer is at a height, as shown in the figure.
Mass of hammer, m = 1000 kg
Acceleration due to gravity, g = 9.8 m/s²
Height of hammer, h = 10 m
E_p = mgh

Potential Energy = mgh = 1000 × 9.8 × 10 = 98000 J
The hammer has no kinetic energy at the initial position since it is not in motion. Hence total energy = potential energy + kinetic energy
= 98000 + 0 = 98000 J
When the hammer strikes the pile, the potential energy of it changes to kinetic energy. Now let’s consider the top end of the pile reaching the ground level. When the hammer touches the pile at the ground level the velocity with which the hammer hits the pile is the maximum. Hence the kinetic energy of the hammer is also the maximum.
Kinetic energy is \(\frac{1}{2}\) mv²

From the equations of motion, we can write v² = u² + 2as = u² + 2gh (since a = g and s = h)
Initial velocity of hammer, u = 0
Therefore, v² = u² + 2gh = 0 + 2 × 9.8 × 10 = 196
Kinetic energy = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 1000 × 196 = 98000
The total energy of the hammer just before it touches the pile at the ground level + kinetic energy of the hammer just before it touches the pile at the ground level = 0 + 98000 = 98000 J

Question 49.
What conclusion can we draw from comparing the total energy when the hammer is at the top and the total energy at the instant of touching the ground?
Answer:
One form of energy can be converted into another without loss of energy.

Question 50.
What are the similarities between work and energy?
Answer:

• Both work and energy are scalar quantities.
• Unit of work and energy is Joule.

Question 51.
From where do we mainly get the energy needed for our various activities?
Answer:
The main source of energy on the Earth is the Sun. Observe figure.

Question 52.
How do plants use solar energy?
Answer:
Plants absorb solar energy during photosynthesis and store it as chemical energy.

Question 53.
How is the energy we get from food related to the Sun?
Answer:
We get chemical energy stored by plants through food.

Question 54.
Write down the energy change that takes place here.
Answer:
Solar energy is converted to chemical energy.

Question 55.
Is the energy production in power plants related to the Sun?
Answer:
Yes.

Question 56.
How is the energy in petrol and diesel related to the Sun?
Answer:
Petrol, diesel etc are fossil fuels. Fossil fuels are formed when organic matter undergoes chemical changes. We get the same energy from the sun as we do from fossil fuels.
Observe the schematic diagram of a hydroelectric power station.

Question 57.
Which energy is possessed by the water in the reservoir?
Answer:
Potential Energy

Question 58.
How did the water in the reservoir get this potential energy?
Answer:
The water in the water bodies evaporates due to the heat of the sun and rain reaches the higher regions. They get their high potential energy from the sun.

Question 59.
A pipe that carries water downwards is called a penstock; which energy is possessed by the flowing water?
Answer:
Kinetic energy
A generator converts this kinetic energy into electric energy. Reservoir stores a large quantity of water. So it can serve as a huge source of energy.

Question 60.
Complete the energy conversion starting from the energy of water in a hydroelectric power plant to the energy obtained when lighting a bulb in a house.
Answer:

Water in the dam	Water flowing down	When the turbine of the generator rotates	When the generator works	When the bulb glows
Potential energy	Kinetic energy and potential energy	Mechanical energy	Electrical energy	Light energy

Question 61.
Bulbs are not the only devices used in our houses. Name some electric devices.
Answer:
Motor, Refrigerator, Electric iron, Television

Question 62.
Do they all use electric energy at the same rate?
Answer:
No.

Question 63.
Using a stopwatch, find the time required to place all ten books on the table and record it on the table.

Answer:

Question 64.
Is the quantity of work done by all the children the same?
Answer:
Yes.

Question 65.
Which child completed the work first?
Answer:
D

Question 66.
Who has done more work in one second?
Answer:
D

Unit of power = J/s = watt.
The SI unit of power is named watt to commemorate the scientist, James Watt. Its symbol is W.
The unit megawatt and kilowatt are also used while measuring the larger values.
1 kW = 10³W = 1000 W
1 MW = 1,000 kW = 10⁶ W = 10,00,000 W
The power of vehicles and motors is expressed in horsepower also.
1 HP = 746 W

Question 67.
Write down the names of some electric appliances we use in our houses and their marked power.
Answer:
Approximate values of each are given.
Light bulb (2 – 100 watts)
Ceiling fan (20 – 75 watts)
Refrigerator (150 – 800 watts)
Electric Iron (500 – 2000 watts)

Question 68.
Calculate the power of an electric iron if 15 kJ work is done in 20 s.
Answer:
Work = 15 kJ = 15000 J
Time = 20 s
Power = Work / time
= 15000/20
= 750 W