# NCERT Solutions Class 9 Science Chapter 11 Work and Energy

## NCERT Solutions Class 9 Science Chapter 11 Work and Energy

NCERT Solutions Class 9 Science Chapter 11 Work and Energy: National Council of Educational Research and Training Class 9 Science Chapter 11 Solutions – Work and Energy. NCERT Solutions Class 9 Science Chapter 11 PDF Download.

### NCERT Solutions Class 9 Science Chapter 11: Overview

 Board NCERT Class 9 Subject Science Chapter 11 Chapter Name Work and Energy Topic Exercise Solutions

### NCERT Solutions Class 9 Science Chapter 11 – Work and Energy

Part 1

1.) A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:  Given, force = 7 N, The displacement is along the direction of applied force.

Displacement = 8 m.

As we know that,

Work done = force × displacement

Work done = 7 ×8

Work done = 56 J

The value of Work done by applied force is 56 J.

Part 2

1.) When do we say that work is done?

Answer: When an object get displaced because of applied force then we say that work is done. There are three conditions for Work done

• The value of force is not zero.
• The value of displacement is not zero.
• The angle between force and displacement is not right angle.

If these conditions are satisfied then we say that work is done.

2.) Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer: Let, the magnitude of force F is acted on the particle and it get displacement.  The magnitude of displacement is d. Then work done is defined as,

Work done = force × displacement

W = F ×d.

This is the expression of work done.

3.) Define 1 J of work.

Answer: Joule is the unit of work done as well as energy. 1 J  is defined as, when an object get displaced of 1 m because of force of 1 N applied along the direction of displacement.

As we know that,

Work done = force × displacement

1 J = 1 N × 1 m

4.) A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Answer: A pair of bullocks exerts force of 140 N.

Force = 140 N, The field being ploughed is 15 m because of applied force by pair of bullocks.

Displacement = 15 m.

As we know that,

Work done = force × displacement

Work done  = 15 × 140

Work done = 2100 J

The work done by pair of bullocks is 2100 J.

Part 3

1.) What is the kinetic energy of an object?

Answer:  Kinetic energy related with velocity of particle.

Definition – The energy possesses in the body because of it’s velocity is called kinetic energy.

SI unit of kinetic energy is J.

2.) Write an expression for the kinetic energy of an object.

Answer: Consider an object has mass ‘m’ kg is moving with velocity ‘v’ m/s then the kinetic energy of that object is,

Kinetic energy = ½ mv²     ……        (1)

Also we write this equation in the form of momentum.

As we know that,

momentum = p =  mv     ……         (2)

Multiply and divide equation 1 by m then it becomes,

Kinetic energy = ½ (mv)²/m

Put the value of momentum from equation 2 in above equation,

Kinetic energy  = (p²)/2m     ……    (3)

Where,

P – momentum,

m- mass of the object,

v- velocity.

Equation 1 and equation 3 are gives us the value of kinetic energy.

3.) The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocityis doubled? What will be its kinetic energy when its velocityis increased three times?

Answer: Given, Mass = m, Velocity = 5 m/s, kinetic energy = 25 J.

Firstly we calculate mass of the object,

As we know,

Kinetic energy = ½ mv²

25 = ½ × m × 5²

25 × 2 = 25 × m

m = 2 kg.

Now we calculate energy when velocity becomes doubled.

Let new velocity = v’ .

v’ = 2v

v’ = 2 × 5

v’ = 10 m/s.

As we know,

Kinetic energy = ½ mv²

Kinetic energy = ½ ×2× 10²

Kinetic energy = 100 J.

If velocity becomes doubled the its kinetics energy increases by 4 time.

Now we calculate energy when velocity increases by 4 times. Let new velocity = v”.

v” = 3v.

v” = 3 × 5

v” = 15 m/s.

As

Kinetic energy = ½ mv²

Kinetic energy = ½ × 2 × 15²

Kinetic energy = 225.

If the velocity increases by 3 times then Kinetic energy increases by 9 times.

Trick- If the mass of object remains constant then Kinetic energy ∝  v².

The kinetic energy increases with square of velocity.

Part 4

1.)  What is power?

Answer: Power is the quantity depend on rate of work done.

Definition- Rate of doing work is called power.

If an object do work done of ‘W’ in ‘t’ seconds then power is,

Power = w/t.

Power is the ratio of work done and time.

We can define power in terms of energy also.

Power is defined as a rate of transfer of energy.

Power is the ratio of energy and time.

Power = energy/time.

2.) Define 1 watt of power.

Answer: Let the object do work done of 1 J in 1 seconds then we say that power is 1 Watt.

3.) A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Answer: Given, energy = 1000 J, time = 10 s.

As we know that,

Power = energy / time

Power = 1000/10

Power = 100 watt.

The value of power is 100 watt.

4.) Define average power.

In total journey of work done, the rate of transfer of energy is different so the quantity average power is most important.

Average power- Rate of doing work at different rates with different time interval is called average power.

Average power = total work done/total time.

### Exercises Solution

1.) Look at the activities listed below. Reason out whether ornot work isdone in the light of yourunderstanding of theterm ‘work’.

• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A wind-mill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.

Answer: we know that workdone is depend on

1.) Applied force

2.) Displacement

3.) Direction of applied force and displacement

• Suma is swimming in a pond.

In this example, Suma applied force on water. The water displaced in backward direction. Thus workdone is positive.

• A donkey is carrying a load on its back.

In this example,  donkey exerts force in upward direction. The displacement is also done. But the direction of force and displacement is perpendicular to each other. Thus, workdone in this case is zero.

• A wind-mill is lifting water from a well.

Windmill exerts force in upward direction while lifting water from a well. Displacement of water is also in upward direction.  Thus workdone will be positive.

• A green plant is carrying out photosynthesis.

In this example, value of force is zero thus workdone becomes zero.

• An engine is pulling a train.

The direction of force and displacement is same . Thus workdone becomes positive.

• Food grains are getting dried in the sun.

The value of displacement in this case is zero. Thus workdone becomes zero.

• A sailboat is moving due to wind energy.

Sailboat exerts force in force forward direction. The direction of displacement of sailboat is also foreword. Thus workdone becomes zero.

2.) An object thrown at a certain angle to the ground moves inacurved path and falls back to the ground. The initial and thefinal points of the path of the object lie on the samehorizontalline. What is the work done by the force of gravity on the object?

Answer: As we know that, conditions for  workdone-

• The applied force must not zero.
• The displacement of particles also not be zero.
• The angle between force and displacement does not 90.

In above case, initial and final position of the object is at same point and thus displacement becomes zero. There fore we say work done must be zero.

3.) A battery lights a bulb. Describe the energy changesinvolvedin the process.

Answer:  A battery lights bulb. Energy changes in this example as,

Chemical energy in a battery converts into electrical energy and electrical energy converts into light energy.

4.) Certain force acting on a 20 kg mass changes its velocityfrom5 m/s to 2 m/s. Calculate the work done by the force.

Answer: Given, mass = 20 kg, initial velocity= u = 5 m/s,

Final velocity= v = 2 m/s.

As we know that, work done is also defined as change in kinetic energy.

Workdone = final kinetic energy – initial kinetic energy.

Workdone = ½ mv² – ½ mu²

Workdone = ½ m (v² -u²)

Workdone= ½ × 20 (2²-5²)

Workdone = 10 (4-25)

Workdone =10 × (-21)

Workdone = -210 J

The workdone in this process will be 210 J.

5.) A mass of 10 kg is at a point A on a table. It is moved to apointB. If the line joining A and B is horizontal, what is the workdone on the object by the gravitational force? Explain youranswer.

Answer: Given, mass = 10 kg,

As we know that, gravitational force exerts on the object in downward direction only. The displacement by gravitational force are always vertically downwards.

But in above case, the particle did not move up or down hence vertical displacement becomes zero.

And we know that, if value of displacement is zero then workdone also zero.

The value of workdone by gravitational force in above example is zero.

6.) The potential energy of a freely falling object decreasesprogressively. Does this violate the law of conservation ofenergy? Why?

Answer: Law of conservation of energy states that total energy of the system remains constant.

If the potential energy decreases then Kinetic energy increases or vice versa but there is no change in total energy of that system.

In above case, the objects falls under influence of gravitational force. As we know that, potential energy is depend on height of the object. Thus potential energy of freely falling object decreases progressively but the velocity of the object increases continuously.  So there will increase in kinetic energy. So this statement does not violate the law of conservation of energy.

7.) What are the various energy transformations that occurwhenyou are riding a bicycle?

Answer: While riding a bicycle, muscular energy is converted into mechanical energy, Kinetic energy,  heat energy produced by friction between tyre and road.

8.) Does the transfer of energy take place when you push ahuge rock with all your might and fail to move it? Where isthe energy you spend going?

Answer: No, there no transfer of energy while pushing a stationary rock but rock remains in stationary state. All the muscular energy is converted into heat energy and our body heat up.

9.) A certain household has consumed 250 units of energyduringa month. How much energy is this in joules?

Answer: As we know that,  household unit of energy is unit.

1 unit = 1 kilowatt hour.

Given, 250 unit = 250 × kilowatt hour

But we know that,

1 kilo watt= 10³ watt  and

1 hour  = 3600 seconds

250 units = 250 × 10³ × 3600

250 units = 900 × 10⁶ joule

350 units = 9 × 10⁸ J.

250 units means 9 × 10⁸ J of energy.

10.) An object of mass 40 kg is raised to a height of 5 m abovetheground. What is its potential energy? If the object is allowed tofall, find its kinetic energy when it is half-way down.

Answer: Given, Mass = 40 kg,

Height = 5 m,  g = 10 m/s².

As we know that, potential energy is depend on height,

Potential energy = mgh

Potential energy = 40 × 10 × 5

Potential energy = 2000 J.

The potential energy at the height 5 m is 2000 j.

When objects falls then its potential energy decreases and kinetic energy increases. At mid point, the object possesses half potential energy and half kinetic energy.

Thus kinetic energy at half way down = ½× potential energy

Kinetic energy at half way down = ½ × 2000

Kinetic energy at half way down = 1000 J

The kinetic energy at half way down will be 1000 J.

11.) What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer: We know that, workdone is depend on

• Applied force
• Displacement
• Angle between force and displacement

The direction between direction of gravitational force and displacement of the satellite is 90⁰. Thus the workdone vanishes.

The workdone by gravitational force on satellite is zero.

12.) Can there be displacement of an object in the absence of anyforce acting on it? Think. Discuss this question with yourfriends and teacher.

Answer: yes, there be displacement of an object ij the absence of any force acting on it.

We studied first law of motion. The force on an object moving with constant velocity is zero. So there will be a displacement in absence of external force also.

Example – The stone moving in space with constant velocity.

14.) An electric heater is rated 1500 W. How much energydoes ituse in 10 hours?

Answer: Given, power = 1500 W. Time = 10 h.

As we know that,

Power = energy/ time

Energy = power × time

Energy = 1500 × 10

Energy = 15000 watt hour.

Energy = 15 × 10³ watt hour

Energy = 15 kilo watt hour.

The energy consumed by that heater is 15 kilowatt hour.

16.) An object of mass, m is moving with a constant velocity,v. How much work should be done on the object in order to bringthe object to rest?

Answer: Initially the object is in motion.

Given, Initial velocity = v m/s

Mass of the object = m kg.

Thus initial kinetic energy = ½ mv².

We have to calculate Workdone on the object in order to bring the object into rest.

So final velocity becomes zero. Thus final value of kinetic energy becomes zero.

We know that, Change in kinetic energy is called as workdone.

Workdone = change in kinetic energy.

Workdone = ½ mv² -0

Workdone = ½ mv².

The workdone required on the object in order to bring in rest is ½ mv².

17.) Calculate the work required to be done to stop a car of 1500 kgmoving at a velocity of 60 km/h?

Answer: Given, mass of car = 1500 kg, initial velocity = 60 km/h = 60 ×5/18 = 16.67 m/s.

Finally it stop, so final velocity= v = 0 m/s.

We know that,

workdone = change in kinetic energy.

Workdone = final Kinetic energy- initial kinetic energy

Workdone = ½ mv² – ½ mu²

Workdone = ½ m (v² -u²)

Workdone = ½ × 1500 (0² – 16.67²)

Workdone = 750 (-277.89)

Workdone = 208417.5 J

The workdone to stop that car is 20.8 × 10⁴ J.

18.) In each of the following a force, F is acting on an object ofmass, m. The direction of displacement is from west to eastshown by the longer arrow. Observe the diagrams carefullyand state whether the work done by the force is negative,positive or zero.

1. In first case, the direction of force is perpendicular to displacement. So the value of workdone is zero.
2. In second case, the direction of displacement is along applied force . Thus work done in this case must be positive.
3. In third case, direction of displacement is opposite to applied force. Thus the workdone must be negative.

19.) Soni says that the acceleration in an object could be zeroeven when several forces are acting on it. Do you agree withher? Why?

Answer: Yes, I am agree with soni’s statement.

The multiple forces acting on abject in different direction. If the resultant of them is zero then acceleration must be zero.

For example,

Four forces acting on the object at different direction as shown in figure. The resultant of them is zero. Thus they produced zero acceleration.

20.) Find the energy in kW h consumed in 10 hours by fourdevices of power 500 W each.

Answer: Given, time = 10 hours,

Power = 500 W.

But we know that,  1 watt = 10‐3 kilowatt.

Power = 500 w = 500 × 10‐3 kilowatt.

Power = 0.5 kilowatt

Power = 500

As we know that,

Power = energy/ time

Energy = power × time

Energy = 0.5 × 10

Energy = 5 kilowatt h.

Each device consumes 5 kilowatt energy.

So total energy consumed by 4 devices = 4 × 5

Total energy consumed by 4 devices = 20 kilowatt h.

Total energy consumed by 4 devices are 20 kilowatt h.

21.) A freely falling object eventually stops on reaching theground. What happens to its kinetic energy?

Answer: As we know that, motion of free fall. When the object is at maximum height then it possesses potential energy only. If we dropped the object from that height then, it’s velocity increases continuously. Thus kinetic energy increases and potential energy decreases at every seconds.

When the object reached at ground then potential energy becomes zero and Kinetic energy becomes maximum.

Updated: September 9, 2021 — 6:55 pm