PSEB Solutions for Class 9 Science Chapter 11 Work, Energy and Power

PSEB Solutions for Class 9 Science Chapter 11 Work, Energy and Power

PSEB 9th Class Science Solutions Chapter 11 Work, Energy and Power

→ All living beings need food for their energy needs.

→ Machines too need fuels like petrol and diesel for energy.

→ In our daily life, we consider any useful physical or mental activity as work.

→ Work done by the force acting on a body is equal to the product of the magnitude of the force and the displacement in the direction of the force.

→ When a force acts in a direction opposite to the direction of displacement produced, then the work done is negative.

→ When a force acts in the direction of displacement then the work done is positive.

→ For us, the sun is the major natural source of energy.

→ In addition to this, we can also get energy from nuclei of atoms, from the interior of earth, and tides.

→ If a body has the capacity to do work then it is said to possess energy.

→ The object which does work loses energy and the object on which work is done gains energy.

→ The sum total of potential and kinetic energy of the body is called mechanical energy.

→ The energy present in the body due to its motion is called it’s kinetic energy.

→ The kinetic energy of an object increases with the increase of its velocity.

→ When an object is raised to some height above the ground, the work done against gravitational force is stored in the object as potential energy.

→ We can transform one form of energy into another form.

→ The rate of doing work or transformation of energy is called power.

→ The unit of energy is joule but this unit is small. Its bigger unit is kilowatt hour (kWh) 1 kWh = 3.6 × 106 J

→ HP Energy consumed in industries and commercial establishments are expressed in kilowatt-hour which is also called unit (B.O.T. unit)

→ In order to survive all living beings have to do some basic activities for which energy is required. This energy they get from the food they eat.

→ According to the law of conservation of energy, energy can neither be created nor destroyed but it can only be transformed from one form to another form total energy before and after transformation always remains constant.

→ Energy is present in different forms as for example kinetic energy, potential energy, heat energy, chemical energy, electric energy, and light energy.

→ From a scientific point of view for doing work two conditions must be satisfied:

• some force must act on the object
• the object must be displaced in the direction of the force.

→ The unit of power is the watt.

→ 1 kilowatt = 1000 watt.

→ 1 watt is the power of that agent which does 1 joule of work in 1 second.

→ Energy: The capacity of doing work is called energy.

→ Kinetic Energy: The capacity of the bodies to do work due to the motion present in them is called kinetic energy.

→ Potential Energy: The capacity of an object to do work clue to change in its position or configuration is called its potential energy.

→ Law of Conservation of Energy: Energy can neither be created nor destroyed but it can be changed from one form into another form.

→ Joule: If one Newton (N) of force acts on a body of 1 kg and displaces it through 1 m then work done on the body is one joule.

→ Power: The rate at which energy is supplied or consumed is called power. The unit of power is the watt (W).
Power =  Energy Supplied / Time 

→ Watt: If a source supplies or consumes energy at the rate of 1 Joule (J) per second then the power of the source, is said to be one watt.

→ Work: If the force acting on a body displaces the body in the direction of force then work is said to be done by the force.
Work = Force × Displacement
i.e. W = F × S

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Long Answer Type Questions:

Question 1.
What is potential energy? Deduce an expression for the gravitational potential energy of a body of mass’m’ placed at height ‘h’ above the ground. Give some practical examples.
Answer:
Potential Energy: It is the energy possessed by an object when it is raised above or below the surface of earth.
Mathematical Expression: Consider a body of mass ‘m’ raised to a height ‘h’ above the surface of earth to a position ‘B’. In order to lift the body to this height we have to apply minimum force equal to mg, the weight of the body in the upward direction from position ‘A’.

∴ Work done on the body against the force of gravity (W) = Force × Displacement
W = F × S
But F = mg and S = h
W = mg × h
= mgh

The work done (W = mgh) on the body is against the force of gravity
∴ E_p = W = mgh
Since this work done is against the force of gravity, so this energy is called gravitational potential energy. This potential energy does not depend on the path along which the body is moved.

Practical Examples:
1. In olden days there were no digital watches and the watches used to be wound up. When its spring is open then the watch does not work. When its spring is wound up then it starts working.

As is shown in Fig. on winding the spring closes and potential energy is stored up. This potential energy due to change in size and shape of the spring helps to move the hands of watch.
2. Try to open a spring with your both hands. Work is, therefore, being done by your hands. By doing this the spring extends in its length and as such potential energy is stored in it.

Question 2.
What is kinetic energy? Derive a mathematical expression for kinetic energy.
Answer:
Kinetic Energy: It is defined as the energy possessed by a body due to velocity, if body is not in motion then it does not possess kinetic energy.
Mathematical Expression: Suppose a football of mass ‘m’ is at rest and force ‘F’ is applied. Due to the force acting on it, the football covers a distance ‘S’ in’t’ seconds and attains a velocity υ. Thus, acceleration produced in the football is ‘a’

Work done on the football W = F × S ………..(1)
According to Newton’s second law of motion,
F = m × a …………. (2)

Equation (8) proves that the work done on the football is stored in it as kinetic energy.
∴ kinetic energy stored = Work done = 1/2 mv²

Question 3.
What is law of conservation of energy? Explain the law with the help of an example and prove its reality.
Answer:
Law of conservation of energy: According to this law total energy always remain constant or total energy always remains the same. Though one form of energy can be transformed to some other form of energy but total energy still remains constant.

Explanation with example: Throw a ball vertically upwards from the surface of earth. You do some work on the ball. This work gets stored in the form of potential energy. This is called gravitational potential energy. As the ball moves up and up its velocity continue to decrease and its kinetic energy continue to decrease and potential energy goes on increasing. When ball reaches the maximum height its kinetic energy becomes minimum i.e. zero but potential energy becomes maximum. We can prove mathematically that at any position in the path of motion of the ball the sum of potential energy and kinetic energy remains same.

Mathematical Proof for law of conservation of energy:
Consider a ball of mass 10 kg situated at a point 30 m above the surface of earth i.e. point A. Let it be dropped from point A or shown in fig.
At point A
potential energy of ball (P.E.) = m × g × h
= 10 × 10 × 30
= 3000 J
Since, ball is at rest initially thus, at point A it kinetic energy (K.E.) = 0
∴ Total mechanical energy of ball = P.E. + K.E.
= 3000 + 0 = 3000 J ………….(i)

At point B
Ball is above the surface of earth by 20 m.
potential energy (P.E.) = m × g × h
= 10 × 10 × 20
= 2000 J
Using υ² – u² = 2gh
υ² = u² + 2gh
υ² = 0 + 2 × 10 × 10
∴ υ² = 200
∴ Kinetic energy at point B (K.E.) = 1/2 mυ²
= 1/2 × 10 × (200)
= 1000 J
∴ Total mechanical energy of ball at point B = potential energy + kinetic energy
= 2000 + 1000
= 3000 J. ……………… (ii)
Similarly, at point C and D total mechanical energy will be 3000 J.
These equation prove that total energy is always conserved.

Question 4.
If the force acting on the object is not in the direction of motion then how will you consider the work done? Explain giving example and also tell when will the work done be minimum and when it will be maximum?
Answer:
When the force acting on the object is not in the direction of motion. A gardener pushes the lawn mower in the forward direction. He applies force F on the handle HH’ facing the ground.

As is clear from the figure, the gardener is not applying force in the horizontal direction but applies force in a direction making angle θ with the direction of force. In such situation the force acting on the machine drives the roller in the horizontal direction from A to B.
Here the effective force is F cos θ and not F and the vertical component F sin θ balances the weight of the lawn mower.
∴ Work done by the lawnmower (W) = Component of Force × Displacement
= F cos θ × S
1. When the force acts along the direction of displacement then θ = 0° and cos θ = cos 0° = 1
∴ W = F cos θ × S
= F × 1 × S
W = F × S
This time the work done will be maximum.

2. When the force acting on the object is in a direction perpendicular to the direction of motion then θ = 90° and cos θ = cos 90° = 0
∴ W = F cos θ x S
= F cos 90° x S
= F x 0 x S
⇒ W = 0
This time the work done is minimum.

Short Answer Type Questions:

Question 1.
What is work? How can you calculate it? Also give the unit of work.
Answer:
Work: Work done by a force or by an object in the product of applied force and the displacement taking place in the direction of applied force.
If F = force acting on the body
S = displacement of the body in the direction of force
Work done by force, W = F × S ……(i)
we know, when a force acts on a body acceleration is produced in the body.
Thus, if m = mass of body
a = acceleration produced in the body then from Newton’s second law of motion
F = m × a ….(ii)
Using (i) and (ii),
W = m × a × S ………….(iii)

Unit of work: When force is measured in Newton (N) and displacement in metre (m), the
work = Newton × Metre
W = N × m
W = Joule (J)
∴ S.I. unit of work is Joule (J) and C.G.S. unit is erg.
1 Joule = 10⁷ erg.

Question 2.
Show by giving an example that if force acting in the body does not produce any displacement then the work done will be zero?
Answer:
This statement can be understood by the following example.
If a child tries to push a car by applying his, maximum force but the car does not displace even through 1 centimeter, then in the language of physics we can say that the child has done no work.
In this situation suppose the child applies force F and displacement S = 0 then
∴ Work done by the child, W = F × S
= F × 0
⇒ W = 0

Question 3.
A stone tied to one end of string is moved in a circle. How much work is done by the centripetal force in this circular motion?
Answer:
A stone is made to move in a circle as shown the fig. then the finger with which you hold the string will experience some force. The force acting on the moving stone is known as centripetal force.

This force acts radially inwards towards the centre of the circle. If in this situation the stone gets detached then it will fly off in the direction tangential along AT or BT as shown in fig. Now, the centripetal force acts perpendicular to the direction of motion, thus no work is done by the centripetal force.

Question 4.
What is power? Write its SI unit also.
Answer:
Power. Harish and Karan climbed up a tree to pluck 60 mangoes each. They started climbing at the same moment of time. Harish got his 60 mangoes in 30 minutes whereas Karan got his 60 mangoes in 60 minutes. That means Karan took more time to do the same work. Both did the same work, both had same energy but their power was not same i.e. Power of Harish is more than that of Karan.

Question 5.
Prove by an experiment that mechanical energy can be transformed into heat energy.
Or
When we hammer a nail into a wooden block the nail gets heated up? Why?
Answer:
Place a nail on a wooden block and hammer it, some part of the nail goes into the wooden block, when nail is completely into the block due to hammering, when nail in further hammered then nail, hammer and block all gets heated up. When hammer is lifted up for hammering, its potential energy increases due to its position.

When it falls on the nail then its whole energy is transformed into kinetic energy of the nail and the nail goes into the wooden block. When nail is completely embedded into the block then mechanical energy of the hammer converts into heat energy of the nail, block and hammer meaning thereby that mechanical energy gets converted or transformed into heat energy.

Question 6.
Differentiate between Potential Energy and Kinetic Energy.
Answer:
Difference between Potential energy (P.E.) and Kinetic energy (K.E.)

Potential Energy	Kinetic Energy
1. The potential energy of an object depends upon its position and size.	The kinetic energy of an object is due to its motion or velocity.
2. Potential energy (P.E.) = m × g × h	Kinetic energy (K.E.) = 1/2 mυ2
3. Potential energy of an object depends upon its height above the ground or its depth below the ground surface.	The kinetic energy of an object depends upon its velocity.

Question 7.
A horse and a dog are running with the same velocity. If the mass of horse is ten times the mass of the dog then what will be ratio of their kinetic energy?
Solution:
Suppose the mass of dog = m
∴ Mass of the horse = 10m
Velocity of horse = Velocity of dog = υ (say)

Question 8.
Two masses m and 2m are dropped from height h and 2h. On reaching the ground, which will have a greater kinetic energy and why?
Answer:
K.E. of mass m = P.E. lost by mass m = mgh
K.E. of mass 2m = P.E. lost by mass 2m
= 2m × g × 2h = 4mgh
Hence, mass 2 m will have a greater kinetic energy on reaching the ground.

Question 9.
Two objects having same mass’m’ are moving with velocities υ and 2υ. Find ratio of their kinetic energies.
Answer:
Let bodies A and B each have same mass m and their respective velocities are υ and 2υ

Question 10.
In what conditions the physical capacity of a man to do work decreases?
Answer:
After sickness and in old age the physical capacity of a man to do work decreases because the energy of his body muscles becomes less.

Question 11.
Explain clearly the difference between energy and power with the help of an example.
Answer:

• Energy: Energy is the ability to do work and is equal to the magnitude of total work which could be done by the energy. It has no relation with the time.
• Power: It is the rate of doing work. It has no relation with the magnitude of total work.
• Example: A worker taken one hour to complete a piece of work, whereas the other worker takes 2 hours to complete the same piece of work. In this case, both the workers are doing the same work i.e. both are consuming same amount of energy.

But first worker did the same work in half the time as compared to other worker thus. First worker has doubled the power than the other worker.

Important Formulae:

Numerical Problems (Solved):

Question 1.
J of energy is applied to lift a box of mass 0.5 kg. How much high it would be raised?
Solution:
Potential energy (P.E.) = 1 J
Mass of the box (m) = 0.5 kg
Acceleration due to gravity, (g) = 10 m/s²
Height to which box is raised, (h) = ?
We know, potential energy (P.E.) = m × g × h
1 = 0.5 × 10 × h
1 = 5 × h
or h = 1/5 = 0.2
∴ Height to which box would be raised, (h) = 0.2 m

Question 2.
A woman pulls water-filled bucket weighing 5 kg from a well 10 m deep in 10 s. What is her power?
Solution:
Mass of water filled bucket, (m) = 5 kg
Depth of well(h) = 10 m
Acceleration due to gravity, (g) = 10 m s^-2
Work done by woman(W) = P.E
= m × g × h
= 5 × 10 × 10
= 500 J
Time (t) = 10 s

Question 3.
A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How much potential energy does he gain? Take r = 9.8 m/s²?
Solution:
Here, m = 50kg, g = 9.8 m/s², h = 100m
Work done by the body = mgh
= 50 × 9.8 × 100
= 49000 J
= 4.9 × 10⁴ J
Gain in P.E. = Work done = 4.9 × 10⁴ J

Question 4.
A man drops a 10 kg rock from the top of a 20 m ladder. What will be its kinetic energy when it reaches the ground? What will be its speed just before it hits the ground? Does the speed depend on the mass of the rock? (Take g = 10 m s^-2)
Solution:
Here u = 0, m = 10 kg, h = 20 m, g = 10 m s^-2
We know, υ² – u² = 2gh
∴ υ^-2 – 0^-2 = 2 × 10 × 20
or υ = √400
= 20 m s^-1
K.E. = 1/2 mυ²
= 1/2 × 10 × (20)²
= 2000 J
Speed does not depends on the mass of the rock because the acceleration due to gravity under which the rock falls does not depend on mass.

Question 5.
A rocket of 3 × 10⁶ kg mass takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. Calculate the potential energy, and the kinetic energy. (Take g = 10 m s^-2).
Solution:

Question 6.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Solution:
Here, P = 1000 W
= 1 kW
Total time, t = 2 × 28 hours
= 56 hours
Total energy consumed = P × t
= 1kW × 56h
= 56kWh
∴ Cost of 1 kWh = ₹ 3.00
∴ Cost of using electricity for Feb = 3 × 56
= ₹ 168.

Question 7.
The power of a motor pump is 5 kW. How much water per minute the pump an raise to height of 20 m? Take g = 10 ms^-2.
Solution:
Here, P = 5 kW = 5000 W, t = 1 min = 60s, h = 20 m, g = 10 ms^-2

Question 8.
Calculate the electricity bill amount for the month of November of a family if 4 tube lights of 40 W each for seven hours, a TV of 150 W for three hours and two bulbs of 60 W each for four hours are used per day. The cost per unit is ₹ 3.50.
Solution:
Power of each tube light = 40 W
∴ Total power of 4 tube lights = 4 × 40 W
= 160 W
Energy consumed by 4 tube lights each day
= 160 W × 7 h
= 1120 Wh
= 1120/1000 = 1.112kWh
Energy consumed by T.V. per day = 150W × 3h
= 450 Wh
= 450/1000 = 0.45 kWh
Energy consumed by 2 bulbs per day = 60W × 4h × 2
= 480 Wh
= 480/1000 = 0.48 kWh
Total energy consumed per day by all appliances = 1.12 + 0.45 + 0.48 = 2.05 kWh
Total energy consumed in 30 days = 2.05 kWh × 30
= 61.50 kWh
Cost of 1 kWh = ₹ 3.50
Cost of 61.50 kWh = 3.50 × 61.50
= ₹ 215.25

Question 9.
A person carrying 10 bricks each of man 2.5 kg. on his head moves to a height 20 metres in 50 seconds. Calculate the power spent in carrying bricks of the person. (g = 10 ms^-2)
Solution:
Total mass (m) = 10 × 2.5 kg
= 25 kg.
Time (t) = 50 s
Displacement (S) = 20m
g = 10 ms^-1
Force exerted by person (F) = mg
= 25kg × 10 ms^-2
= 250 N
Work done = Force × Displacement
= F × S
= 250 N × 20 m
= 5000 J

Question 10.
A car of 1000 kg moving with a velocity of 30 m/s stops with uniform acceleration after covering a distance of 50 m on application of brakes. Find the force applied by the brakes on the car and also work done.
Solution:
u = 30 ms^-1
υ = 0
S = 50m
Now υ² – u² = 2aS
(0)² – (30)² = 2 × a × 50
– (30 × 30) = 100 × a
a = –900/100
∴ a = – 9m/s²
Force, F = m × a
= 1000 × 9
= 9000 N
work done by the brakes, W = F × S
= 9000 × 50
= 450000 N-m
= 4.5 x 10⁵ J

Question 11.
A freely falling hammer of mass 1 kg 7 falls on a nail fixed in a block of wood. If the hammer falls from a height of 1 m then what will be the kinetic energy just before striking the nail? (Take g = 10 m s^-2)
Solution:
Mass of the hammer, (m) = 1kg
Height of the hammer (h) = 1 m
Acceleration due to gravity (g) = 10ms^-2
using υ² – u² = 2gS
υ² – (0)² = 2 × 10 × 1
υ² = 20 ………..(i)
Now kinetic energy of hammer (K.E.) = 1/2 × m × υ²
= 1/2 × 1 × 20 [From (i)]
= 10 J

Question 12.
A car is moving with a speed of 54 km/h. What will be the kinetic energy of the boy of mass 40 kg sitting in the car?
Solution:
Speed of the boy = speed of the car
= 54 km/h
= 54 × 5/18 m s^-1
= 3 × 5 m s^-1
= 15 m s^-1
Mass of the boy (m) = 40kg
∴ kinetic energy (K.E.) of the boy = 1/2 mυ²
= 1/2 × 40 × (15)²
= 20 × 15 × 15
= 4500 J

Question 13.
1 Joule of energy is used for one heart beat. Calculate the power of the heart if it throbs 72 times in one minute.
Solution:
Work done in 1 heart beat = 1 J
∴ Total work done by heart in 72 beats = 72 × 1 J
= 72 J
Time (t) = 1 min
= 1 × 60 s
= 60 s

Very Short Answer Type Questions:

Question 1.
State the relation between commercial unit of energy and joules.
Answer:
1 commercial unit of energy (or 1 kWh) = 3.6 × 10⁶ Joule.

Question 2.
How much work is done on a body of mass 1 kg whirling on a circular path of radius 5m?
Answer:
Work done is zero.

Question 3.
What is the SI unit of power?
Answer:
The SI unit of power is Watt.

Question 4.
A ball is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the maximum height?
Answer:
The kinetic energy of the body changes into its potential energy.

Question 5.
If the heart works 60 joules in one minute, what is its power?
Answer:

Question 6.
Name the term used for the sum of kinetic energy and potential energy of a body.
Answer:
Mechanical energy.

Question 7.
How many joules make one-kilowatt hour?
Answer:
1 kilowatt-hour = 3.6 × 10⁶ J.

Question 8.
What should be the change in velocity of a body required to increase its kinetic? energy to four times of its initial value?
Answer:
The velocity of the body should doubled at constant mass.

Question 9.
Under what conditions the work done by a force is zero inspite of displacement being taking place?
Answer:
When displacement is in a direction perpendicular to the applied force.

Question 10.
What is the power of a machine which does 2000 joules of work in 10 seconds?
Solution:

Question 11.
What is the SI unit of kinetic energy?
Answer:
Joule.

Question 12.
Water flows down the mountains to the plains. What happens to the potential energy of water?
Answer:
Potential energy of water will decrease. It will change to kinetic energy of water.

Science Guide for Class 9 PSEB Work, Energy and Power InText Questions and Answers

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

Here force (F) = 7 N
Displacement (S) = 8m
∴ Work done (W) = ?
We know, W = F × S
= 7N × 8m
= 56N – m
= 56 J

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacment.
Answer:
When the displacement of the object is in the direction of force then,
Work done (W) = Force (F) × Displacement (S)

Question 3.
Define 1 J of work.
Answer:
Joule. Work done on the body is said to be 1 J if IN of the force acting on it displaces the body in its own direction through a distance of 1 m.
We know, W = F × S
or 1 J = 1 N × 1 m
= 1N – m

Question 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 ?
Solution:
Force exerted by bullocks (F) = 140 N
Length of the field ploughed (S) = 15 m
Work done in ploughing the field (W) = ?
We know, W = F × S
W = 140 N × 15 m
= 2100 N – m
= 2100 J

Question 5.
What is kinetic energy of an object?
Answer:
Kinetic Energy: Kinetic energy of an object is the quantity of motion possessed by it.
Kinetic energy of an object of mass ‘m moving with velocity υ is = 1/2 mυ²
S.I. unit of kinetic energy is Joule
Examples:

1. Stone in motion
2. Blowing wind
3. Rotating wheel

Question 6.
Write an expression for the kinetic energy of an object.
Answer:
Kinetic energy of an object of mass ‘m’ moving with velocity ‘υ’ is
E_k = 1/2 × mass × (velocity)²
= 1/2 × m × (υ )²
= 1/2mυ²
∴ Expression for kinetic energy (E_k) = 1/2 mυ²

Question 7.
The kinetic energy of an object of mass’m’ moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled ? What will be its kinetic energy when its velocity is increased to three times.
Solution:
Given, Mass of an object = m
Velocity of the object (v) = 5 ms^-1
Kinetic energy of the object (Efc) = 25 J

Question 8.
What is Power ?
Answer:
Power: The rate of doing work or the rate of transformation of energy is called power. Its S.I. unit is watt (W) or Joule/sec (J s^-1)
If an agent does work /W’ in time ‘t’ then
Power =  Work done / Time taken to do work 
or P =  W/t 

Question 9.
Define 1 watt of Power.
Answer:
Watt. It is the power of that agent or machine which can work at the rate of 1 Joule in 1 second.

Question 10.
A lamp consumes 1,000 J of electric energy in 10 s. What is its power ?
Solution:
Here, W = 1000 J
Time (t) = 10 s
Power (P) = ?

Question 11.
Define Average Power.
Answer:
Average Power: It is defined as the ratio of total energy consumed to the total time taken.
∴ Average Power (P_av) =  Total energy consumed / Total time taken 

PSEB 9th Class Science Guide Work, Energy and Power Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A wind-mill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the sun.
7. A sail boat is moving due to wind energy.

Answer:

1. Suma while swimming is applying her muscular force in a particular direction and gets displaced. Therefore, work is being done by Suma.
2. The load being carried by donkey is acting in the downward direction perpendicular to the horizontal direction of displacement. And when the force acts perpendicular to the direction of displacement then no work is done. Therefore, donkey is not doing any work.
3. Work is being done because in lifting water, the displacement, as well as force, are in vertically upward direction.
4. A green plant carrying photosynthesis does no work since neither there is force applied nor any displacement in direction of force applied.
5. An engine pulling a train is doing work since displacement is in direction of force applied.
6. No work is done on food grain. However part of heat suplied coverts moisture of grains into steam which rises up increasing P.E.
7. Work is being done since force and displacement is there in the same direction.

Question 2.
An object 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 object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
The work done by the force of gravity will be zero. This is because the displacement is in a horizontal direction while the force is acting vertically downward perpendicular to this direction of displacement.

In this situation θ = 90°
∴ cos θ = cos 90° = 0
Now work done (W) = F cos θ × S
= F × 0 × S
W = 0

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
Chemical energy of the chemicals in the battery is first being converted to electric energy. Then the electric energy of the battery is converted into heat energy and light energy by the bulb.

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 m s^-1 to 2 ms^-1. Calculate the work done by the force.
Solution:
Here mass (m) = 20 kg
Initial velocity (u) = 5ms^-1
Final velocity (v) = 2ms^-1

∴ Negative sign shows that there is decrease in velocity due to opposing force which is doing work.

Question 5.
A mass of 10 kg is at a point Aon atable. It is moved to a point B. If line joining A and B is horizontal, what is die work done on fee object by gravitational force ? Explain your answer.
Solution:

An object of mass 10 kg is displaced in the horizontal direction from point A to point B but the gravitational force is acting vertically downward which makes an angle of 90° with the direction of displacement.
∴ Work done by the gravitational force (W) = F cos θ × S
= F cos 90° × S
= F × 0 × S
= 0

Question 6.
The potential energy of a freely falling object decreases progressively. Does this Violate the law of conservation of energy? Why?
Answer:

• It does not violate the law of conservation of energy. When the height of freely falling body decreases, its potential energy decreases but kinetic energy increases.
• Kinetic energy increases by the same amount as potential energy has decreased.
• At any time the sum of kinetic energy and potential energy remains conserved.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle ?
Answer:
When we are riding a bicycle and pedalling it, the energy of our muscles gets transformed into heat energy and kinetic energy. This kinetic energy is used in doing work against the frictional energy offered by the road.

Question 8.
Does the transfer of energy takes place when you push a huge rock with all your might and fail to move it ? Where is the energy you spent going?
Answer:
Although you have not been able to move the heavy rock, you are very much tired and this has reduced your energy. Since we have failed to move the heavy rock, work appears to be zero.

While pushing the stone, you had to stretch your muscles, heart had to pump more blood and in making these changes, your energy is definitely lost. The work done by you on your body is not zero. You may have to eat some food to compensate for the work done by your muscles and heart.

Question 9.
A certain household has consumed 250 units of electric energy during a month. How much energy is this in joules ?
Solution:
We know, 1 unit of energy = 1 kilowatt hour (1 kWh)
= 1 kW × 1 h
= (1 × 1000 watt) × (1 × 60 × 60 s)
= 36 × 10⁵ J
= 3.6 × 10⁶ J
∴ 250 units of energy = 250 × 3.6 × 10⁶ J
= 900 × 10⁶ J
= 9 × 10⁸ J

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is half-way down. Take g – 10 ms “2.
Solution:
Here mass of the object (m) = 40 kg
Height above the ground (h) = 5m
Acceleration due to gravity (g) = 10 ms^-2
Potential energy of the object at a height of 5 m (E_p) = m × g × h
= 40 × 10 × 5 J
= 2000 J
Let υ be the velocity of the object when it has come halfway down
Distance moved by the object (S) = 5/2 = 2.5 m
using υ² – u² = 2gS
υ² – (0)² = 2 × 10 × 2.5
υ² = 2 × 25
or υ² = 50
Kinetic energy of the object on reaching half way down (E_k) = 1/2 mυ²
= 1/2 × 40 × 50
= 1000 J

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

When a satellite moves around the earth the force of gravity is directed inward along the radius of the circular path while the direction of motion is along the tangent which is perpendicular to the radius. In this way force of gravity and displacement are mutually at right angle to each other as a result of which the work done on the satellite is zero.
We know, work done (W) = F cos θ × S
= F cos 90° × S
= F × 0 × S
= 0

Question 12.
Can there be a displacement of any object in the absence of any external force ?
Answer:
In the absence of any external force the displacement of the object is possible if the object is moving with a uniform velocity. And if the object is in the state of rest then the displacement is not possible in the absence of external force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer.
Answer:
A person holds a bundle of hay over his head for 30 minutes and gets tired but the force of gravity acting on the bundle does not displace the bundle of hay in the direction of force of gravity. Since there is no displacement in the direction of force, therefore, no work is said to be done by him.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours ?
Solution:
Given, Power of the heater (P) = 1500 W
Time for which heater is used (t) = 10 hr
Energy used by the heater in 10 hours (E) = ?
We know, energy used = Power × Time
= 1500 Watt × 10 hrs
= 1500 Wh
= 1500/1000 k Wh
= 15kWh
= 15 units

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest ? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
Energy transformation in oscilated pendulum:
Initially bob is at rest at its mean position, thus in kinetic energy is zero. We can consider its potential energy equal to zero in this position.

When we take the bob of the pendulum to one side its height goes on increasing and we have to do some work against the force of gravity. This work is stored in the bob as its potential energy. Thus, when a bob is released from one of its extreme position, i.e. state of maximum displacement B at this position its kinetic energy is zero and potential energy is maximum.

Now, the bob is in motion and is moving towards its mean position A its height goes on decreasing that means its potential energy also goes on decreasing whereas its velocity goes on increasing and hence, its kinetic energy increases. Because bob is moving through the air, thus, some energy is consumed, against the force of friction due to air. This causes increase in speed of molecules of air and thus, kinetic energy of the molecules increase.

At the mean position, kinetic energy of the bob becomes maximum and potential energy becomes minimum. Due to inertia of motion bob does not stop here but it moves to the other side of its mean position. Its height again starts increasing so that potential energy also increases, but kinetic energy continues to decrease. When bob reaches at the extreme position ‘O’ its potential energy becomes maximum and kinetic energy becomes zero.

Bob does not stop here it comes back towards its mean position ‘A’. At every point of its motion, sum of kinetic energy and potential energy of the bob along with energy of air molecules remains constant. Thus, during oscillation of the bob of a simple pendulum total energy remains conserved.

Amplitude of the simple pendulum depends on the total energy of the bob. The energy transfered to the molecules of the air by the oscillating bob can never be recovered. Thus total energy of the bob goes on decreasing. When, the bob transfers whole of its energy to the molecules of the air then its total energy becomes zero and it comes to rest at its mean position. Thus there is no violation of the law of conservation of energy.

Question 16.
An object of mass ‘m’ is moving with velocity V. How much work should be done on the object in order to bring the object at rest?
Solution:
Let an object of mass m be initially moving with velocity υ and finally be brought to rest by the application of an opposing force F after covering a distance S.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km h^-1.
Solution:

Question 18.
In each of the following force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force in negative, positive or zero

Answer:

1. In Fig. (a) displacement is in a direction perpendicular to direction of force applied, therefore work done is zero. There is no displacement in direction of force.
2. In Fig. (b), the work done is positive since force and displacement are in the same direction.
3. In Fig. (c), the work done is negative since displacement is in a direction opposite to direction of force.

Question 19.
Soni says that acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Yes, acceleration can be zero when a large number of forces are acting on a body and their resultant is zero.
We kno w, a =  F/m 
=  0/m 
∴ a = 0
Illustration:

1. If two equal and opposite forces are acting on an object, acceleration of the object is zero.
2. If three forces are simultaneously acting on an object and can be represented in magnitude and direction by three sides of the triangle in the same order, the body is in equilibrium and will have zero acceleration even when three forces are acting on it.
3. Even with more than 3 forces acting on a body, it could have zero acceleration if the resultant of all force is zero.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Solution:
Power of 1 device = 500 W
Total power of 4 devices = 4 × 500 W = 2000 W
Time for which 4 devices used (t) = 10 hrs
Energy consumed in 10 hours = Power × Time
= P × t
= 2000W × 10 h
= 20000 Wh
= 20000/1000 kWh
= 20 kWh

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
A freely falling body on reaching the ground finally stops. Its kinetic energy gets transformed into other forms of energy such as heat, sound and light etc. and then into its potential energy.

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