# Class 9 Science Chapter 11 Important Questions

Class 9 Science Chapter 11 Important Questions of Work and Energy. These questions cover the entire chapter 11 of grade 9 science from NCERT Book. Class 9 Science Chapter 11 Extra Question Answers contains all the practice questions which are important for school exams.

## Class 9 Science Chapter 11 Important Questions for 2020-2021

#### Class 9 Science Chapter 11 Important Extra Questions Set – 1

##### List two conditions which need to be satisfied for the work to be done on an object.

W = Fs
Work is done when
(a) a force acts on an object
(b) object is displaced.

##### Explain the following terms with one example each: (a) Positive work (b) Zero Work

(a) When force acts in the direction of motion of body work done is positive. When a lawn roller is pulled forward, work done is positive.
(b) When force is perpendicular to the direction of motion, work done is zero. A porter carrying load does no work.

##### From where do we get energy for the life process?

The energy required for the various life process comes from the food that we eat.

Zero.

##### Moment of force and work done by a force have the same units. Then, what is the difference between them?

Moment of a force produces rotatory motion in a body whereas when a forced does work on a body it produces translatory motion in it.

###### When an arrow is short from its bow, it has kinetic energy. From where does it get this kinetic energy?

A stretched bow possess potential energy on account of a change in its shape. When the arrow is released, the potential energy of the bow gets converted into the kinetic energy of the arrow.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 2

##### A coolie is walking on a railway platform with a load of 27kg on his head. What is the amount of work done by him?

Work done by the coolie is zero, as W = Fs cos 900 = 0.

##### What is the amount of work done by a main in pressing a rigid wall with a force of 400N?

Zero because there is no displacement.

##### Which law was verified experimentally by James Perscott Joule?

Law of conservation of energy.

##### State the law of conservation of energy.

The law of conservation of energy states that energy can neither be created, nor destroyed, it can only be transformed from one form to another.

##### Name the two common forms of mechanical energy.

Kinetic energy and potential energy.

###### When an electric bulb is switched on, what energy transformation takes place?

Electrical energy changes into light energy and heat energy.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 3

##### By what factor does the kinetic energy of a body increases when its speed is doubled?

By a factor of 4 (KE ∝ v2).

##### What is negative work.

Work done against friction is a negative work.

##### Out of a light and a heavy body having equal kinetic energy, which one will is move fast?

The lighter body because v = √(2KE/m)

##### What type of energy is stored in spring of a watch?

Elastic potential energy.

##### Can kinetic energy of a body be negative?

No, because both m and v2 are always positive.

##### Give two examples from everyday life where work is done.
• We apply a force to lift a book at a height and the book rises.
• When a bullock pulls a cart, it moves as work is done by the bullock on the cart.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 4

##### An electrical cells converts which forms of energy into which other form?

Chemical energy gets converted into electrical energy

Electric motor.

##### What kind of energy transformation takes places at thermal power station?

Heat energy is converted into electrical energy.

##### A car is accelerated on a levelled road and attains a velocity four time its initial velocity. In this process how does the potential energy of the car change?

Potential energy (= mgh) does not change as it does not depend on velocity.

1800.

##### On what factors the work on a body depends?

The work done on a body depends upon two factors:

1. Magnitude of the force (F) and
2. The displacement through which the body moves (s)

#### Class 9 Science Chapter 11 Important Extra Questions Set – 5

##### A student picks up four books from the floor, walks across the room through some distance with the books at the same height and then keeps these books at the new place. In which of these sequence of actions work is said to be performed?

When the student is picking up books from the floor in the beginning.

##### (a) What kind of energy transformation takes places when a body is dropped from a certain height? (b) What is the commercial unit of energy?

(a) When a body falls, its potential energy gradually gets converted into kinetic energy. On reaching the ground, the whole of the potential energy of the body gets converted into kinetic energy.
(b) The commercial unit of energy is kilowatt hour [kWh]
1kWh is the energy used in one hour at the rate of 1000Js-1.

##### What is the relationship between the commercial unit and SI unit of energy?

We know that
1kWh = 1kW × 1h
= 10000 W × 36000s
= 1000 Js-1 × 36000s
= 36000000J
1kWh = 3.6 × 106 J

##### Calculate the work done against the gravity.

Suppose a body of mass m is lifted vertically upwards through a distance h. In this case, the force required to lift the body will be equal to weight of the body, mg (where m is mass and g is acceleration due to gravity), Now,
Work done in lifting a body = Weight of body × Vertical distance
W = mg × h
= mgh
Where W = Work done, and h = Height through which the body is lifted.

##### What is power? How do you differentiate kilowatt from the kilowatt hour?

Power is the rate of doing work. Kilowatt is the unit of power and kilowatt hour is the unit of energy.

##### What is energy?

The capacity of a body to do work is called energy possessed by the body. It is a scaler quantity and is measured in Joule (J).
Generally, for practical purpose a bigger unit called kilojoule (KJ) is used (1 KJ = 1000J).

#### Class 9 Science Chapter 11 Important Extra Questions Set – 6

##### (a) Derive an expression for kinetic energy of an object. (b) If the velocity of an object is doubled. What will be change in its kinetic energy?

Suppose a body of mass m is moving with velocity v. It is brought to rest by applying a retarding force F. Suppose it traverses a distance s before coming to rest.
Kinetic energy of body, KE = Work done by retarding force to stop it.
i.e. Kinetic energy = F.s …. (i)
But Retarding force, F = ma ……(ii)
Initial velocity = v, final velocity = 0
From equation v2 = u2 + 2as, we have
O = v2 – 2as (because here a is retardation)
Distance s = v^2/2a …..(iii)
Substituting value F and s from (ii) and (iii) in (i) we get
Kinetic energy, KE = ma × v^2/2a = 1/2mv2

##### (a) Give one example of each of the following: (i) Small mass but high kinetic energy (ii) Large mass but low kinetic energy (b) Proved mathematically that the total mechanical energy of a freely falling body in air is conserved.

(a) (i) A cricket/hockey ball which has been hit hard and is travelling fast.
(ii) A shot put thrown by an athlete.
(b) Let the body of mass m at height h above the ground starting from rest, be falling freely.
Total energy of the body at height h
= mgh (PE) + 0 (KE) = mgh
After the body has fallen freely through a distance x (say),
KE = mgh, PE = mg (h –x)
Total energy = KE + PE
= mgh + mg (h – x) = mgh
When it reaches the ground KE =1/2 m.2gh = mgh
PE = 0 Total Energy = KE + PE = mgh
Thus the total mechanical energy, which is the sum of KE and PE is always equal to mgh.

##### Two workmen are employed on a building project. (i) Workman 1 drops a hammer, which falls to the ground. The hammer has a mass of 2.0kg and is dropped from a height of 4.8m above the ground. (a) Calculate the change in gravitational potential energy of the hammer when it is dropped. (b) Describe the energy changes from the time the hammer leaves the hand of workman 1 until it is at rest on the ground. (ii) Workman 2 picks up the hammer and takes it back, up the ladder to workman 1. He climbs the first 3.0 m in 5.0 s. His total weight, including the hammer, is 520N. Calculate the useful power which his legs are producing.

(i) Mass, m = 2.0 kg, height h = 4.8 m
Change in gravitational potential energy = mgh
= 2 × 10 × 4.8 = 96 J
(b) Potential energy → Kinetic energy → Heat and/or sound energy
(ii) Weight, mg = 520 N, height h = 3.0 m, time, t = 5.0s
Useful power = (Work done )/(Time taken)
= (mg)h/t = (520×3 )/5
= 312 W

##### Briefly describing the gravitational potential energy, deduce an expression for the gravitational potential energy of a body of mass m placed at a height h, above the ground.

When an object is raised through a certain height above the ground, its energy increases. This is because the work is done on it, against gravity. The energy present in such an object is called gravitational potential energy. Thus the gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity. Consider a body of mass m lying at point P on the earth surface where its potential energy is taken as zero. As weight, mg acts vertically downwards so to lift the body to another position Q at a height h, we have to apply a minimum force which is equal to mg in the upward direction. Thus work is done on the body against the gravity.
We know that,
Work done, W = Fs ….(i)
As F = mg and s = h
Putting these value in equation (i), we get
W = mg × h = mgh
This work done on the body is equal to the gain in energy of the body. This is the potential energy of the body.
Potential energy PE = mgh

##### Show that when body is dropped from a certain height, the sum of its kinetic energy at any instant during its fall is constant.

The mechanical energy (kinetic energy + potential energy) of a freely falling object remains constant. It may be shown by calculation as follows:
Suppose a body of mass m falls from point A, which is at height H from the surface of earth. Initially at point A, kinetic energy is zero and the body has only potential energy.
Total energy of body at point A
= Kinetic energy + Potential energy
= 0 + mgH = mgH …..(i)
Suppose during fall the body is at position B. The body has fallen at a distance x from its initial position. If velocity of body at B is v, then from formula v2 = u2 + 2as we have
v2 = 0 + 2gx = 2gx
Kinetic energy of body at point B = 1/2 mv2
= 1/2 m × 2gx = mgx
Potential energy of point B = mg (H – x)
Total energy of body at point B = Kinetic energy + Potential energy
= mgx + mg(H – x) = mgH ……(ii)
Now suppose the body is at point C, just above the surface of earth(i.e. just about to strike the earth) Its potential energy is zero.
The height by which the body falls = H
If v is velocity of body at C, then from formula
v2 = u2 + 2as
We have u = 0, a = g, s = H
So, v2 = 0 + 2gH = 2gH
Kinetic energy of body at position C = 1/2 mv2
= 1/2 m × 2gH = mgH
Total energy of body at C
= Kinetic energy + Potential energy
= mgH + 0 = mgH …..(iii)
Thus we can see that the sun of kinetic energy and potential energy of freely body at each point remains constant.
Thus under force of gravity, the total mechanical energy of body remains constant.

##### Uses of Kinetic Energy

(i) The kinetic energy of air is used to run windmills.
(ii) Potential energy due to shape: In a toy car, the wound spring possesses potential energy, and as the spring is released, its potential energy changes into kinetic energy due to which the car moves.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 7

##### A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

Initial velocity = v, then final velocity v’ = 3v
Initial kinetic energy = 1/2 mv2
Final kinetic energy (KE) = 1/2 mv2 = 1/2 m(3v)2 = 9( 1/2 mv2 )
(KE)initial : (KE)final = 1: 9.

##### The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increasing in the kinetic energy of the body is equal to the work done by the force on the body.

v2 – u2 = 2as
This gives, s = (v2 – u2 )/2as
F = ma
We can write work done (W) by this force F
W = Fs
W = ma ((v2 – u2 )/2as)
= 1/2 m (v^2-u^2 ) = 1/2 mv2 – 1/2 mu2
= Final KE – Initial KE

##### A light and a heavy object have the same momentum, find out the ratio of their kinetic energies. Which one has a large kinetic energy?

Linear momentum of first object P1 = m1v1 and of second object p2 = m2v2
But p1 = p2
or, m1v1 = m2v2
If m1 < m2 then v1 > v2
(K.E.)1 = 1/2 (m1v1) v1 = 1/2 p1v1
and (K.E) 2 = 1/2 (m2v2)v2 = 1/2 p2v2
So, ((KE)_1)/((K.E)_2 ) = (1/2 p1v1)/(1/2 p2v2 ) = v_1/v_2
But v1 > v2
Therefore, (K.E)1 > (K.E)2

##### Four men lift a 250 kg box to a height of 1m and hold it without raising or lowering it. (a) How much work is done by the mean in lifting the box? (b) How much work do they do in just holding it? (c) Why do they get tired while holding it? (g = 10ms-2)

(a) F = 250kg × 10ms-2 (g = 10ms-2)
= 2500N
s = 1 m
W = F.s
= 2500 N × 1m
= 2500 Nm = 2500J
(b) Zero as the box does not move at all while holding it.
(c) In order to hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force, muscular effort is involved. So, they feel tired.

##### A car of mass 900kg is travelling at a steady speed of 30m/s against a resistive force of 2000N. (i) Calculate the kinetic energy of the car. (ii) Calculate the energy used in 1.0s against the resistive force. (iii) What is the minimum power that the car engine has to deliver to the wheels? (iv) What form of energy is in the fuel, used by the engine to drive the car?

(i) Kinetic energy = 1/2 mv2
= 1/2 × 900 × (30)2 = 1/2 × 900 × 900
= 4,05,000J
(ii) Energy used = Work done against resistive force
= Force × Distance
= 2,000 × 30 = 60,000J = 60kJ
(iii) Minimum power = (Energy used )/(Time taken )
= 60,000J/1s = 60,000W = 60Kw
(iv) Chemical energy.

##### Question: 1

A body thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Work done is zero. This is because equal and opposite work is done in the two paths.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 8

##### A force of 20N displace a body through a distance of 1m at an angle of 600 from its own direction. Calculate the amount of work done.

Here, force F = 20N, displacement, s = 1m
Angle between force and displacement ∅ = 600.
Work done, W = Fs cos ∅ = 20 × 1 × cos 600 = 20 × 1×1/2 = 10J.

##### Calculate the work done in lifting 200kg of a mass through a vertical distance of 6m. Assume g = 10m/s2.

Work in lifting the mass is done against gravity.
Therefore, the work done is W = mgh
We know, m = 200kg, g = 10m/s2 and h = 6m
W = mgh
= 200 × 10 × 6
= 12000J

##### Find the velocity of a body of mass 100g having kinetic energy of 20J.

Here mass of the body m = 100g = 0.1 kg
Kinetic energy, K.E. = 20J = 20Nm = 20 kg
Kinetic energy = 1/2 mv2
v = √((2×kinetic energy )/(m ))
= √((2×20kg m^2/s^2 )/(0.1kg ))
= 20m/s
So the velocity of the body is 20m/s.

##### The speed of a vehicle of mass 500kg increases from 36km/h to 72km/h. Calculate the increase in kinetic energy.

Given mass, m = 500kg
The given unit of speed is km/h. It is converted into m/s
1 km/h = (1 × 1000meter )/(3600 second ) = (5 )/( 18 )m/s
Initial speed u, = 36km/h = 36 × (5 )/( 18 )m/s = 10m/s
Final Speed, v = 72 km/h = 72 × (5 )/( 18 )m/s = 20m/s
Gain in KE = Final KE – Initial KE
= 1/2 mv2 – 1/2 mv2 = 1/2 m × (v2 – u2)
= 1/2 × 500 × [(20)2 – (10)2]
= 1/2 × 500[400 – 100]
= 1/2 × 500× 300
= 75000joule
= 7.5 × 104 J

##### A man of mass 60kg runs up a fight of 30 steps in 15 seconds. If each step is 20cm high, calculate the power developed by the man. (Taking g = 10m/s2).

Height of each step = 20cm = 0.20cm
Height of the 30 steps, H = 30 × 0.20
Time t = 15 seconds
Force exerted by man against gravity
F = Weight of man = mg
Work done by man = Force × Height = mg × H
Power P = (Work )/(Time ) = (mgH )/(t )
= (60 ×10 ×6.0 )/(15 ) = 240Joule.

##### Question: 2

You lift a heavily packed carton of mass m in vertically upwards direction through a height h. What is the work done (i) by you on the carton, (ii) by force of gravity on the carton?

(i) Work done by me is positive and having a value = mgh. This is because I am applying force vertically upward direction on the carton to hold it and displacement is also in the same direction.
(ii) Work done by the force of gravity on the cartoon = -mgh. This is because force is vertically downwards but motion is vertically upward.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 9

##### A machine raises a load of 750 N through a height of 16m in 5 seconds. Calculate the power which the machine works.

Given force, F = 750N, displacement = height h = 16m and time t = 5
Work done by machine,
W = Force × Displacement = F × h
= 750N × 16 Joules
Power of machine, P = W/t = (750N×16J )/(5s ) = 2400 watt.

##### If the power of a motor is 40kW at what speed can it raise a load of 20,000N?

Given power, P = 40kW = 40 × 103 = 40, 000W
Force applied, F = Load = 20,000N
If v is the speed of load, then we have
P = Fv
Speed v = P/F = 40,000/20,000 = 2m/s

##### A engine can pump 30,000 litres of water to a vertical height of 45 metres in 10 minutes. Calculate the work done by the machine and its power.

Volume of water raised = 30,000 litres = 30,000/1000 m^3 = 30 m3.
Mass of water raised, M = Volume × Density
= (30m3) × (103 kg/m3)
= 30 × 103 kg
Height, h = 45 meter
Work done by machine,
W = Weight of water raised × Height
= (Mg) × h (30 × 103 × 9.8) × 45
W = 1.323 × 107 joule
Time taken, t = 10 minutes = 10 × 60 = 600 seconds.
Power, P = W/(t ) = (1.323×〖10〗^7 J )/(600 s)
= 22 × 103W
= 22kW.

##### A boy is moving on a straight road against a friction force of 5N. After travelling a distance of 1.5 km he forgot the correct path at a round about of radius 100m. However he moves on the circular path for one and half cycle and then he moves forward up to 2.0 km. Calculate the work done by him.

Force F = 5 ; s = 1500m + 2 × 100m + 200m
Work done, W = F.s
W= 5 × [1500 + 200 +2000]
= 18500J.

##### The power of a motor pump is 2kW. How much water per minutes can the pump raise to a height of 10m? (Given g = 10ms-1).

P = W/t = mgh/t
= (m×10 ×10 )/60 = 2000W
m = 12000/10
= 1200kg.

##### Question: 3

Anil is doing work at a rapid rate but works for only one hour. Ashok does work at some what slower rate but continues to work for six hours. Who has the greater power? Who has more energy?

Anil has greater power because his rate of doing work is more. Ashok has more energy as he worked for a longer time and the total work done by him in definitely more.

#### Class 9 Science Chapter 11 Important Extra Questions Set – 10

##### If an electric iron of 1200W is used for 30 minutes every day, find the electric energy consumed in the month of April.

Power P = 1200/1000 = 1.2 kW
Time t, = 30/60 = 0.5h
Electric energy, E = Power × time × days
= 1.2 × 0.5 × 30
= 18kWh.

##### . An automobile engine propels a 1000kg car along a levelled road at a speed of 36kmH-1. Find the power is the opposing frictional force is 100N, suppose after travelling a distance of 200m, this car collides with another stationary car B of same mass and comes to rest. Let its engine also stop at the same time. Now car B starts moving on the same level road without getting its engine started. Find the speed of car B just after the collision.

An automobile engine propels a 1000kg car along a levelled road at a speed of 36kmH-1. Find the power is the opposing frictional force is 100N, suppose after
ma = mb = 1000kg.v = 36km/h = 10m/s
Frictional force = 100N
Since that car A moves with a uniform speed it means that the engine of car applies a force equal to the frictional force.
Power = (Force ×Distance )/(time ) = F.v
= 100N × 10m/s = 1000W
After collision
ma ua + mb ub = ma va + mb vb
1000 × 10 + 1000 × 0 = 1000 × 10 + 1000 ×vb
vb = 10 ms-1

##### Give three examples when the object is not displaced in spite of a force acting on it

Three examples when force being supplied and still there is no displacement are:

1. A man pushing a stationary truck.
2. A man pushing a rigid wall.
3. A boy carrying a basket on his head and standing still. Here, force of gravity acts on the basket, but there is no displacement.

What types of energy transformation takes places in the following:

1. Electric heater
2. Solar battery
3. Dynamo
4. Steam engine and
5. Hydroelectric power station?