Lakhmir Singh Manjit Kaur Class 9 Physics 4th Chapter “Work and Energy” solution

Lakhmir Singh Manjit Kaur Class 9 Physics 4th Chapter “Work and Energy” solution

Lakhmir Singh and Manjit Kaur Physics solution: “Work and Energy” Chapter 4. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Physics solution Chapter 4 . Here we have given Chapter 4 all solution of Class 9. Its help you to complete your homework.

  • Board – CBSE
  • Text Book Physics
  • Class – 9
  • Chapter – 04

Lakhmir Singh Manjit Kaur Class 9 Physics 4th Chapter Solution (First Part)

Very Short Answer Type Questions:

1.) How much work is done when a body of mass m is raised to a height h above the ground ?

Answer: Given, Mass = m, height = h

We did a workdone against gravitational potential energy so

Workdone = potential energy= m×g×h.

Where g- acceleration due to gravity.

2.) State the SI unit of work.

Answer: SI unit of work is Joule.

3.) Is work a scalar or a vector quantity ?

Answer: Work  has only magnitude  so it is a scalar quantity.

4.) Define 1 joule of work.

Answer: If a body get displacement of 1 m due to 1 N force exert on it then that workdone is 1 Joule.

5.) What is the condition for a force to do work on a body ?

Answer: Condition for workdone-.The body must get displacement due to applied force on it.

6.) Is energy a vector quantity ?

Answer: No, energy is scalar quantity.

7.) What are the units of (a) work, and (b) energy ?

Answer: Work and energy has same dimensions therefore they have same units.

  • Unit of work- Joule.
  • Unit of energy- Joule.

8.) What is the work done against gravity when a body is moved horizontally along a frictionless surface ?

Answer: Workdone against gravity is depend on height. But in this example height is zero thus work also zero.

9.) By how much will the kinetic energy of a body increase if its speed is doubled ?

Answer: The relationship between speed and kinetic energy is

Kinetic energy ∝ v2

If speed become doubled then kinetic energy increases by 4 times.

10.) Write an expression for the kinetic energy of a body of mass m moving with a velocity v.

Answer: If a body of mass m moving with velocity v then it’s Kinetic energy is,

Kinetic energy = ½ mv²

11.) If the speed of a body is halved, what will be the change in its kinetic energy ?

Answer: As we know the relation between speed and kinetic energy,

Kinetic energy ∝ v2

If the speed halved then it’s kinetic energy reduces four times.

12.) On what factors does the kinetic energy of a body depend ?

Answer: Kinetic energy depends on speed and mass of the object.

13.) Which would have a greater effect on the kinetic energy of an object : doubling the mass or doubling the velocity ?

Answer: Kinetic energy is depend on mass and square of speed. So doubling velocity will effect greater on kinetic energy.

14.) How fast should a man of 50 kg run so that his kinetic energy be 625 J ?

Answer: Given, Mass = 50 kg, kinetic energy = 625 J.

As we know,

Kinetic energy =  ½ mv²

625 = ½ ×50 × v²

625 × 2/50 = v²

25 = v²

Taking square root on both side,

v = 5 m/s

That man must move with 5 m/s.

15.) State whether the following objects possess kinetic energy, potential energy, or both :

(a) A man climbing a hill

(b) A flying aeroplane

(c) A bird running on the ground

(d) A ceiling fan in the off position

(e) A stretched spring lying on the ground.

Answer:

(a)  A man climbing a hill- that man possess kinetic as well as potential energy.

(b) A flying aeroplane- the aeroplane possess kinetic as well as potential energy.

(c) A bird running on the ground- Due to motion, bird possess only kinetic energy.

(d) A ceiling fan in the off position- Fan does not have translation motion so it possess potential energy.

(e) A stretched spring lying on the ground- Due to specific position of spring, it contains only potential energy.

16.) Two bodies A and B of equal masses are kept at heights of h and 2h respectively. What will be the ratio of their potential energies ?

Answer: As we know,

Potential energy = m × g × h.

(Potential energy of A / potential energy of B) = (m × g × h)/(m × g × 2h)

= 1/2 .

The ratio of Potential energy of A to potential energy of B is 1:2.

17.) What is the kinetic energy of a body of mass 1 kg moving with a speed of 2 m/s ?

Answer: Given, Mass = 1 kg, speed = v =2m/s.

As we know,

Kinetic energy =  ½ mv²

Kinetic energy = ½ × 1 × 2

Kinetic energy = 1 J.

Kinetic energy of that body is 1 J.

18.) Is potential energy a vector or a scalar quantity ?

Answer: Potential energy has only magnitude so it is a scalar quantity.

19.) A load of 100 kg is pulled up by 5 m. Calculate the work done. (g= 9.8 m/s2)

Answer: Given, Mass = 100 kg, it pulled up so h = 5 m, g= 9.8 m/s2

If the workdone is against gravitational force then that workdone is equal to gravitational potential energy.

Potential energy = m × g × h.

Potential energy = 100 × 9.8 × 5

Potential energy = 4900 J.

Thus, the value of workdone  is 4900 J.

20.) State whether the following statement is true or false :

The potential energy of a body of mass 1 kg kept at a height of 1 m is 1 J.

Answer:  False,

We verify this as,

Given, Mass = 1kg, height = 1 m, and workdone = 1 J,

1 J = m × g × h.

1 J = 1× 9.8 × 1

1 J = 9.8 J.

Left hand side is not equal to right hand side. Thus the statement is false.

21.) What happens to the potential energy of a body when its height is doubled ?

Answer: Potential energy is directly proportional to height.

When height is doubled then it’s potential energy also becomes double.

22.) What kind of energy is possessed by the following ?

(a) A stone kept on roof-top.

(b) A running car.

(c) Water stored in the reservoir of a dam.

(d) A compressed spring.

(e) A stretched rubber band.

Answer:

(a) A stone kept on roof-top – the stone is at rest so it possess only potential energy.

(b) A running car.- Due to motion of car it possess only kinetic energy.

(c) Water stored in the reservoir of a dam – Due to specific position of water it possess potential energy.

(d) A compressed spring- Due to specific position of stretched spring it possess potential energy.

(e) A stretched rubber band- Due to specific position of stretched rubber band it possess potential energy.

Short Answer Type Questions

24.) What are the quantities on which the amount of work done depends ? How are they related to work ?

Answer: Workdone depends on force and displacement. Also workdone depends on angle between force and displacement. The relationship between force, displacement and workdone is as,

Workdone = f × d × cosø

Where,

f- force,

d- displacement,

ø- angle between force and displacement.

25.) Is it possible that a force is acting on a body but still the work done is zero ? Explain giving one example.

Answer:yes, itis possible that a force acting on a body is non-zero but still the work done is zero.

If the force acting on a body in perpendicular direction then workdone  will zero.

Also if the body does not displaced by force then the workdone also zero.

26.) A boy throws a rubber ball vertically upwards. What type of work, positive or negative, is done :

(a) by the force applied by the boy ?

(b) by the gravitational force of earth ?

Answer: If a boy throws a rubber  vertically upwards then the direction of displacement is also upward.

(a) The force exerted by the boy is in upward direction so workdone by that boy is positive.

(b) The gravitational force always exerts in downward direction but direction of displacement of the ball is upwards hence the workdone is negative.

27.) Write the formula for work done on a body when the body moves at an angle to the direction of force. Give   the meaning of each symbol used.

Answer: Workdone on a body when the body moves at an angle to the direction of force is calculated by,

W= f × d × cosø

Where,

W- workdone

f- force

d- displacement

Ø- angle between force and displacement.

28.) How does the kinetic energy of a moving body depend on its

(i) speed, and (ii) mass ?

Answer: (i) kinetic energy is directly proportional to square of speed of the body.

(ii) Kinetic energy is directly proportional to the mass of moving body.

29.) Give one example each in which a force does (a) positive work (b) negative work, and (c) zero work.

Answer: Examples of

(a) Positive work- If four men push a car and car displaced in the direction of force then the workdone by that men is a positive work.

(b) negative work- If a stone is thrown in vertically upward then the workdone by gravitational force is negative.

(c) zero work- The workdone by gravitational force exerted by sun on earth is zero because angle between force and displacement is 90⁰.

30.) A ball of mass 200 g falls from a height of 5 metres. What is its kinetic energy when it just reaches the ground ? (S = 9.8 m/s2).

Answer: Given, Mass = m =200 g.

Height = h = 5 m.

31.) Find the momentum of a body of mass 100 g having a kinetic energy of 20 J.

Answer: Mass = m = 100 g

m = 0.1 Kg,

kinetic energy = 20 J.

as we know that

kinetic energy =1/2  m × v²

Multiply both side by mass,

m × Kinetic energy  =1/2 m × m × v²

m × Kinetic energy  =1/2 (m × v)²

But m × v = momentum = p

m × kinetic energy =1/2p²

p² =2× m × kinetic energy

p²= 2×0.1 × 20

p²  =  4

Taking square root on both side,

p = 2 kg m/s

The momentum of that body is 2 kg m/s.

32.) Two objects having equal masses are moving with uniform velocities of 2 m/s and 6 m/s respectively. Calculate the ratio of their kinetic energies.

Answer: Given velocity of first object = 2 m/s,

velocity of second object is = 6 m/s,

Mass of first object = mass of second object = m.

We know the formula of Kinetic energy.

Kinetic energy = ½ mv²

The ratio of kinetic energy = (kinetic energy of 1st object)/(kinetic energy of 2nd object)

= (1/2 × m × 2² )/(1/2 × m × 6² )

= 4/36

= 1/9

The ratio of kinetic energy of first object to second object is 1:9.

33.) A body of 2 kg falls from rest. What will be its kinetic energy during the fall at the end of2 s ? (Assume g = 10 m/s2)

Answer: Given, mass= 2 kg, time = 2 second,

It falls from rest so initial velocity =0 m/ss.

We find final velocity using first equation of kinematics.

v = u + a × t

v = 0 + 10× 2

v = 20 m/s

As we know,

Kinetic energy = ½ m× v²

Kinetic energy= ½ × 2 × 20²

Kinetic energy = 400 J.

The value of kinetic energy at the end will be 400 J.

34.) On a level road, a scooterist applies brakes to slow down from a speed of 10 m/s to 5 m/s. If the mass of the scooterist and the scooter be 150 kg, calculate the work done by the brakes. (Neglect air resistance and friction)

Answer: Initial speed = u = 10 m/s,

Final speed = v = 5 m/s, total mass= 150 kg.

As we know,

Workdone = (final kinetic energy) – ( initial kinetic energy)

Workdone= (1/2m × v²)- (1/2 m× u²)

Workdone= ½ × m ×( v²-  u²)

Workdone = ½ × 150 × (5²- 10²)

Workdone = 75 × (25-100)

Workdone = 75 × (-75)

Workdone = – 5625 J.

Minus sign indicates that the workdone is opposite to displacement.

The workdone by breaks is 5625 J.

35.) A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground ? What is its kinetic energy when it reaches the ground ? ( g = 10 m/s2)

Answer: Mass of rock = 10 kg,

Height means displacement covered by rock = x = 5 m.

The rock is dropped so initial speed =0 m/s.

We can calculate final speed by using 3rd equation of kinematics.

v² = u² + 2 ax

v² = 0²+ 2 × 10 × 5

v² = 100

taking square root on both side,

v = 10 m/s

Final velocity at end will be 10 m/s.

Final kinetic energy = ½ m × v²

Kinetic energy = ½ ×10 × 10²

Kinetic energy = 500 J.

The final speed of rock will 10 m/s and kinetic energy will be 500 J.

36.) Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/ s to 10 m/s ?

Answer: Given, Mass = m= 1000 kg,

Initial speed = u = 20 m/s,

Final speed = 10 m/s

As we know,

Workdone = (final kinetic energy) – ( initial kinetic energy)

Workdone= (1/2 m × v²)- (1/2 m× u²)

Workdone= ½ × m ×( v²-  u²)

Workdone = ½ × 1000 × (10²- 20²)

Workdone = 500× (100-400)

Workdone = 500 × (-300)

Workdone = -150000 J.

Workdone = -150 × 10³ J

Workdone = -150 K J.

Negative sign indicates that workdone opposes displacement.

The work done is 150 K J.

37.) A body of mass 100 kg is lifted up by 10 m. Find:

(i) the amount of work done.

(it) potential energy of the body at that height (value of g = 10 m/s2).

Answer: Given, Mass = m = 100 kg,

Height = h = 10 m.

(i) The value of workdone is equal to gravitational potential energy

Gravitational potential energy = m× g×h

Potential energy  = 100 × 10 × 10

Potential energy = 10000

Potential energy = 10 × 10³ J

Potential energy = 10 KJ

Therefore work is 10 KJ.

(ii) If the object gained some height then gravitational potential energy is store into workdone.

Therefore the work is also 10 kJ.

38.) 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 ? (g = 9.8 m/s²).

Answer: Given, mass = 50 kg, Height = 100 m.

As we know that the value of workdone is equal to gravitational potential energy

Gravitational potential energy = m× g×h

Potential energy  = 50× 9.8 × 100

Potential energy = 49000

Potential energy = 49× 10³ J

Potential energy = 49 KJ

The body did word so that energy is stored into gravitational potential energy.  The value of potential energy and workdone is 49 KJ.

39.) When is the work done by a force on a body : (a) positive, (b) negative, and (c) zero ?

Answer: The workdone may be positive, negative and zero depending upon the magnitude and direction of displacement and force.

a) Positive workdone – If an object get displacement in the direction of applied force the workdone will be positive.

b) Negative workdone- If the direction of displacement and applied force is opposite then workdone will be negative .

c) Zero workdone- If any value of force or displacement is zero then workdone also zero. If the angle between force and displacement is 90⁰ then workdone will also zero.

40.) To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 joules ? (g = 9.8 m/s²).

Answer:  Given, Mass = m = 150 kg,

Potential energy = 7350 J,

g = 9.8 m/s².

As we know,

Potential energy = m × g × h

7350 = 150 × 9.8 × h

h = 7350/(150× 9.8)

h = 7350/1470

h = 5 m

The value of height is 5 m.

41.) A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. What will be its potential energy at the end of 2 s? (Assume g = 10 m/s²).

Answer: Given, mass = 2 kg,

Initial velocity = u= 20 m/s, time = 2 seconds,

Finally its velocity becomes zero, v= 0 m/s, Assume g = 10 m/s²

Acceleration due to gravity opposes the  upward motion so we take negative sign for acceleration

There fore g = – 10 m/s².

We can calculate height using second equation of motion

x = ut + ½ a t²

x = 20 × 2 + ½ ×(-10 )× 2²

x = 40 –(5× 4)

x = 40 -20

x = 20 m.

The body reach at height 20 m.

As we know,

Potential energy = m × g × h

Potential energy = 2 × (-10)× 20

Potential energy = -400 J

Potential energy does not negative,

So the value of potential energy at 20 m will be 400 J.

42.) How much work is done when a force of 1 N moves a body through a distance of 1 m in its own direction ?

Answer: Given, Force = 1 N, distance =1 m

We cam calculate work done using the formula,

Work done = force × distance

Work done = 1 × 1

Work done = 1 J.

The value of work done is 1 J

43.) A car is being driven by a force of 2.5 x 1010 N. Travelling at a constant speed of 5 m/ s, it takes 2 minutes to reach a certain place. Calculate the work done.

Answer: Given, Force = 2.5 x 1010 N,

Speed = 5 m/s, time = 2 minutes= 120 seconds, The car is moving with constant speed so acceleration =0 m/s².

We calculate distance covered by car using 2nd equation of motion,

x = ut + ½ at²

x = 5 × 120 + ½ × 0× 120²

x = 600 m.

The car covers displacement of 600 m.

As we know,

Work done = force × displacement

Work done =  2.5 x 1010× 600

Work done  = 15x 1012

Work done  =  1.5 x 1013 J.

The value of work done is   1.5 x 1013 J.

44.) Explain by an example that a body may possess energy even when it is not in motion.

Answer: Sometimes body does not in motion but it possess energy due to its specific position. We can understand this by a simple example.

Example- Water in a dam contains potential energy, stone is placed at top of building contains potential energy.

45.) (a) On what factors does the gravitational potential energy of a body depend ?

(b) Give one example each of a body possessing : (i) kinetic energy, and (ii) potential energy.

Answer: (a) Gravitational potential energy of a body depends on height where the object lie.

(b) (i) Example of a body possessing kinetic energy- water flowing in river possessing kinetic energy, car in a speed possesses kinetic energy.

(ii) Example of a body possessing potential energy- water in a dam possesses potential energy, A stone is placed at certain height possesses gravitational potential energy.

46.) Give two examples where a body possesses both, kinetic energy as well as potential energy.

Answer: Sometimes a body possesses not only kinetic energy but also potential energy.

Examples-

a) A falling stone from certain height. When it is at some height it contains potential energy only but when it reach at ground then it have only potential energy but in middle case it possesses kinetic energy as well as potential energy.

b) Flowing water in river possesses both kinetic energy and potential energy.

47.) How much is the mass of a man if he has to do 2500 joules of work in climbing a tree 5 m tall?

(g = 10 m/ s²)

Answer: Given, The man was at 5 m height so it possesses potential energy = 2500 J,

Height = 5 m, g = 10 m/ s²

Potential energy = m × g × h

2500 = m × 10 × 5

2500 / 50 = m

m = 50 kg

The mass of that man will be 50 kg.

48.) If the work done by a force in moving an object through a distance of 20 cm is 24.2 J, what is the magnitude of the force ?

Answer: Given, distance = 20 cm = 0.2 m,

Work done = 24.2 J,

We know,

Work done = |force| × distance

24.2 = force × 0.2

Force 24.2/0.2

Force = 121 N

The magnitude of force is 121 N.

49.) A boy weighing 40 kg makes a high jump of 1.5 m.

(i) What is his kinetic energy at the highest point?

(ii) What is his potential energy at the highest point? (g = 10 m/s²).

Answer: Given, Mass = m = 40 kg, Height = 1.5 m, g = 10 m/s².

(i) When the boy jumps up then at height point velocity in upward direction becomes zero because of gravitational force.

As we know that, kinetic energy is directly proportional to speed. If value of speed is zero then kinetic energy also zero.

(ii) potential energy = m × g × h.

Potential energy = 40 × 10 × 1.5

Potential energy = 600 J

At height point potential energy becomes zero.

50.) What type of energy is possessed

(a) by the stretched rubber strings of a catapult ?

(b) by the piece of stone which is thrown away on releasing the stretched rubber strings of catapult ?

Answer:

(a) by the stretched rubber strings of a catapult- potential energy.

(b) by the piece of stone which is thrown away on releasing the stretched rubber strings of catapult- kinetic energy as well as potential energy.

51.) Weightlifter is lifting weights of mass 200 kg up to a height of 2 metres. If g = 9.8 m/s², calculate :

(a) potential energy acquired by the weights.

(b) work done by the weightlifter.

Long Answer Type Questions:

52.) (a) Define the term ‘work’. Write the formula for the work done on a body when a force acts on the body in the direction of its displacement. Give the meaning of each symbol which occurs in the formula.

(b) A person of mass 50 kg climbs a tower of height 72 metres. Calculate the work done.

(g = 9.8 m/s²)

Answer: (a) When a body get displacement because of external force acting on it then we say that the work is done.

Work done is scalar quantity. It has dimensions same as energy.

The formula for calculating work done is,

W = F × d

Where,

W- Work done

F – external force

d – displacement due to external force.

(b) Given, Mass = m = 50 kg, height= 72 hours, g = 9.8 m/s².

We know that workdone is stored in gravitational potential energy,

Work done = gravitational potential energy = m × g × h.

Work done = 50 × 9.8 × 72

Work done = 35280 J.

The value of work done is 35280 J.

53.) (a) When do we say that work is done ? Write the formula for the work done by a body in moving up against gravity. Give the meaning of each symbol which occurs in it.

(b) How much work is done when a force of 2 N moves a body through a distance Of 10 cm in the direction of force ?

Answer: (a) When a body displaced due to external force then we say work is done.

It is necessary to work done that body must perform certain displacement also force does not zero.

When a body do work against gravity then the work done is stored in the form of gravitational potential energy and it is calculated as,

W = m × g × h

Where,

W- Workdone

m – mass of the body

g – acceleration due to gravity

h – height.

(b) Given, Force = F = 2 N, distance = d = 10 cm = 0.1 m.

We know that

Work done = force × distance

Work done = 2 × 0.1

Work done = 0.2 J.

The value of work done is 0.2 J.

54.) (a) What happens to the work done when the displacement of a body is at right angles to the direction of force acting on it ? Explain your answer.

(b) A force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface. Calculate the work done if the direction of force is at an angle of 60⁰ to the horizontal surface.

Answer: The value of workdone is depend on applied force, displacement due to force and angle between force and displacement. The value of workdone is calculated bye following formula,

Workdone = f × d × cosø——-1

Where f- force,

d- displacement,

ø- angle between force and displacement.

The angle between force and displacement is right angle means 90⁰ and we know that cos 90 = 0

Put this value in equation 1,

Workdone = f × d × 0

Workdone = 0

If the angle between force and displacement is right angle then value of workdone will be zero.

(b) Given, force = f = 50 N, distance = d = 4 m,

Angle between force and displacement = ø = 60⁰

We know that,

Workdone = f × d × cosø

Workdone = 50 × 4 × cos 60⁰

But cos 60⁰ = ½

Workdone = 50 × 4 × ½

Workdone = 100 J.

The value of workdone is 100 J.

55.) (a) Define the term ‘energy’ of a body. What is the SI unit of energy.

(b) What are the various forms of energy?

(c) Two bodies having equal masses are moving with uniform speeds of v and 2v respectively. Find the ratio of their kinetic energies.

Answer: Energy: (a) Energy is defined as capacity to work. It is a capacity to doing work. Energy is a scalar quantity.

Energy is depend on amount of workdone.

(b) The total energy is called the mechanical energy of the body. It is divided into two types

1) Kinetic energy: The energy possesses by body due to its speed is called kinetic energy.

Example- The energy possesses bye moving car.

2) potential energy: The energy stored in the object due to its specific position is called potential energy.

Example – The energy stored in stone at top of the building.

(c) As we know that, kinetic energy is depend on mass and square of speed.

In this example mass is but speed is different. Let us consider ? Mass of object 1= Mass of object 2 = m,

Velocity of object 1 = v,

Velocity of object 2 = V=2v

(Kinetic energy of body 1)/(kinetic energy of body 2) = (1/2 m×v²)/(1/2× m ×V²)

= v²/(V)²

= v²/(2v)²

= v²/4v²

= ¼.

The ratio of kinetic energy is 1:4.

56.) (a) What do you understand by the kinetic energy of a body ?

(b) A body is thrown vertically upwards. Its velocity goes on decreasing. What happens to its kinetic energy as its velocity becomes zero ?

(c) A horse and a dog are running with the same speed. If the weight of the horse is ten times that of the dog, what is the ratio of their kinetic energies ?

Answer: (a) Kinetic energy: The energy depend on speed of the object is called as kinetic energy. The value of kinetic energy is depend on mass and velocity.

(b) If a body thrown vertically upward then it’s velocity decreases because of gravitational force acting on the body in downward direction. As the speed decreases, the kinetic energy also decreases but potential energy increases with increase in height continuously. When velocity becomes zero that time body gain maximum height. So it have zero kinetic energy and maximum potential energy.

(c) As we know that, kinetic energy is depend on mass and square of velocity.  A horse and a dog are running with the same speed. So the speed is same but masses are different.

Kinetic energy ∝Mass

But mass of horse is ten times greater than mass of dog so kinetic energy horse is ten times greater than kinetic energy of dog.

The ratio of kinetic energy of horse to kinetic energy of dog is 10:1.

57.) (a) Explain by an example what is meant by potential energy. Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.

(b) What is the difference between potential energy and kinetic energy ?

(c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s. Calculate the change in kinetic energy of the ball. State your answer giving proper units.

Answer: (a) Potential energy: The energy stored in the body due to specific position of the object is called potential energy.We understand this concept with the help of an example below.

The stored water in dam contains large amount of energy because of its position.

The formula for calculating gravitational potential energy is,

Gravitational potential energy = m × g ×h.

Where,

m-mass of the object,

g- acceleration due to gravity,

h- height of the object.

(b) Difference between potential energy and kinetic energy.

I) Kinetic energy is defined as the energy contained in the body because of it’s velocity. Potential energy is defined as the energy possesses by body due to it’s specific position.

II) Kinetic energy is depend on velocity but potential energy depends on height or specific position.

III) Formula to calculate Kinetic energy= ½ mv². Formula to calculate potential energy = m g h.

IV) Water in a dam contains potential energy. The flowing water in river possesses kinetic energy.

(c) Mass of the ball  = m = 0.5 kg, Initial velocity = u = 5 m/s, Final velocity = v = 3 m/s.

As we know,

Change in kinetic energy= ½ mu²- ½ mu²

= ½ m (u²-v²)

= ½ ×0.5×(5²-3²)

= 0.25 × (25-9)

= 0.25 × 16

= 4 J.

The change in kinetic energy is 4 J.

58.) (a) What is the difference between gravitational potential energy and elastic potential energy ? Give one example of a body having gravitational potential energy and another having elastic potential energy.

(b) If 784 J of work was done in lifting a 20 kg mass, calculate the height through which it was lifted. (g = 9.8 m/s²)

Answer: (a) Difference between gravitational potential energy and elastic potential energy.

I) Gravitational potential energy is depend on height of the object but elastic potential energy is depend on displacement from the meaning position.

II) At mean position elastic potential energy is zero but at zero height value of gravitational potential energy is zero.

(b) Given, Work done = 784 J, Mass = 20,

g = 9.8 m/s².

As we know that, the work done against gravitational force is stored in gravitational potential energy.

Potential energy = work done  = m × g ×h

Work done = m × g ×h

784 = 20 × 9.8 × h.

h = 784 /(20 × 9.8)

h = 784/ 196

h = 4 m.

The height of lifting is 4 m.

Lakhmir Singh Manjit Kaur Class 9 Physics 4th Chapter Solution (Second Part)

Very Short Answer Type Questions

1.) Name the commercial unit of measurement of energy.

Answer: kilowatt hour is the commercial unit of measurement of energy.

2.) Define kilowatt-hour.

Answer: When an electrical appliance rating one watt is on for one hour then the energy consumed by it is one kilowatt hour.

3.) Name two units of power bigger than watt.

Answer- Units of power bigger than watt are kilo watt, mega-watt and horse power.

1 kilowatt = 10³ watt

1 megawatt = 10⁶ watt.

1 horse power = 746 watt.

4.) Define the term ‘watt’.

Answer: When an electrical appliance consumes energy 1 J per seconds then then the power is called as one watt.

5.) How many watts equal one horse power ?

Answer: One horse power is equal to 746 watt.

1 h.p. = 746 watt.

6.) Name the physical quantity whose unit is watt.

Answer: The unit of power is watt.

7.) What is the power of a body which is doing work at the rate of one joule per second ?

Answer: Given, rate of doing work i.e.

Work done/time = (one joule )/ (one second)

1 watt = (one Joule)/(one second).

The rate of doing work one joule per second is one watt.

8.) A body does 1200 joules of work in 2 minutes. Calculate its power.

Answer: Given, workdone = 1200 J, time = 2 minutes = 120 seconds.

As we know,

Power = workdone/time

Power = 1200/120

Power= 10 Watt.

The value of power is 10 watt.

9.) How many joules are there in one kilowatt-hour?

Answer: As we know,

Kilowatt is the unit of power.

1 kilowatt-hour = 1 × 10³ × watt × 3600 second

= 10³ × 3600 watt – seconds

But, watt = joule/second

= 3.6 × 10⁶( joule /second)/ second

= 3.6 × 10⁶ Joule.

1 kilowatt – hour = 3.6 × 10⁶ J.

10.) Name the quantity whose unit is :

(a) kilowatt

(b) kilowatt-hour

Answer:

(a) Unit of power is kilowatt

(b) Unit of energy is kilowatt hours.

11.) What is the common name of ‘1 kWh’ of electrical energy .

Answer: 1 kWh of electrical energy is 1 unit.

12.) A cell converts one form of energy into another form. Name the two forms.

Answer: A cell converts chemical energy into electrical energy.

13.) Name the device which converts electrical energy into mechanical energy.

Answer: Electrical motor converts electrical energy into mechanical energy. In electrical fan, electrical energy is converted into mechanical energy by using motor.

14.) Name the devices or machines which convert :

(a) Mechanical energy into electrical energy.

(b) Chemical energy into electrical energy.

(c) Electrical energy into heat energy.

(d) Light energy into electrical energy.

(e) Electrical energy into light energy.

Answer:

The devices or machines which convert :

(a) Mechanical energy into electrical energy- Dynamo or electrical generator.

(b) Chemical energy into electrical energy- cell.

(c) Electrical energy into heat energy- electrical heater, oven, iron.

(d) Light energy into electrical energy- solar panel.

(e) Electrical energy into light energy- electrical bulb.

15.) Name the devices or machines which convert :

(i) Electrical energy into sound energy.

(ii) Heat energy into kinetic energy (or mechanical energy).

(iil) Chemical energy into kinetic energy (or mechanical energy).

(iv) Chemical energy into heat energy.

(v) Light energy into heat energy.

Answer:

The devices or machines which converts :

(i) Electrical energy into sound energy- loudspeaker.

(ii) Heat energy into kinetic energy (or mechanical energy)- steam engines.

(iii) Chemical energy into kinetic energy (or mechanical energy)-  Motorcycle.

(iv) Chemical energy into heat energy- LPG gas stove, crackers.

(v) Light energy into heat energy- solar cooker.

Short Answer type Questions

17.) A trolley is pushed along a road with a force of 400 N through a distance of 60 m in 1 minute. Calculate the   power developed.

Answer: Given, force =  F = 400 N, distance=d = 60 m, time = 1 minute= 60 seconds.

We know that,

Power = (workdone)/time

Power = (f × d)/time

Power= (400×60)/60

Power = 400 Watt.

The power developed is 400 watt.

18.) What kind of energy transformations take place at a hydroelectric power station?

Answer: At hydroelectric power station electric current is generated by using gravitational potential energy. The water falling from certain height on dynamo generates electricity. The conversion of energy is as,

Gravitational potential energy > kinetic energy of water > mechanical energy while rotating dynamo > electrical energy.

19.) What kind of energy transformations take place at a coal-based thermal power station?

Answer: The energy transformations at coal based thermal power station is as,

Chemical energy of coal > thermal energy > mechanical energy > electrical energy.

20.) A man weighing 500 N carried a load of 100 N up a flight of stairs 4 m high in 5 seconds. What is the power ?

Answer: Given, the weight of man = 500 N, load = 100 N,

Therefore total force  =  F = 500 N + 100 N

Total force = 600.

Distance = d = 4 m, time = 5 seconds.

We know that,

Power = (workdone)/time

Power = (f × d) / time

Power = (600 × 4) / 5

Power = 480 watt.

The total power is 480 watt.

21.) The power output of an engine is 3 kW. How much work does the engine do in 20 s ?

Answer: Given, power = 3 kW = 3 × 10³ w, time = 20 s,

As we know,

Power = workdone / time

Workdone = power × time

Workdone = 3 × 10³ × 20

Workdone = 60 × 10³ Joule.

Workdone = 60 × kJ

The engine do work of 60 kJ.

22.) An electric heater uses 600 kJ of electrical energy in 5 minutes. Calculate its power rating.

Answer: workdone = 600 kJ= 600 × 10³,

Time = 5 minutes = 300 seconds.

As,

Power = workdone / time

Power = 600 ×10³/ 300

Power = 2 × 10³

Power = 2 kW.

The power rating on heater is 2 kW.

23.) How much electrical energy in joules does a 100 watt lamp consume :

(a) in 1 second ?

(b) in 1 minute ?

Answer: Given, power = 100 watt.

(a) Time = 1 second.

As we know,

Power = energy/time

Energy = power × time

Energy = 100 × 1

Energy = 10 J.

A 100 watt lamp consumes 100 J in one second.

(b) time = 1 minute = 60 seconds

As we know,

Power = energy/time

Energy = power × time

Energy = 100 × 60

Energy = 6000 Joule.

A 100 watt lamp consumes 6000 J in one minute.

24.) Five electric fans of 120 watts each are used for 4 hours. Calculate the electrical energy consumed in kilowatt-hours.

Answer: Given, Power of each fan = 120 watts.

Time = 4 hours. We do not convert time in seconds because we have to calculate energy in kilowatt- hour.

As,

Energy = power × time

Energy = 120 × 4

Energy = 480 watt hour.

Energy consumed by five fans = 5 × 480

Energy consumed by five fans = 2400 watt hour.

Energy consumed by five fans = 2.400 × 10³ watt hour.

Energy consumed by five fans = 2.4 kilowatt hour.

If five electric fans of 120 watts each are used for 4 hours then energy consumed by them is 2.4 kilowatt-hour.

25.) Describe  the energy changes which take place in a radio.

Answer: Cell provides energy to battery. When we on the radio then Chemical energy in cell converted into electrical energy and then sound energy.

Conversion of energy in radio is as,

Chemical energy > electrical energy > sound energy.

26.) Write the energy transformations which take place in an electric bulb (or electric lamp).

Answer: When we on the bulb then electric energy is changes into light energy. If the heater is connected into circuit then, electric energy will convert into heat energy.

27.) Name five appliances or machines which use an electric motor.

Answer: we are almost depend on motors. There are so many instruments in our daily life made up from electric motors. Some examples are as

  1. Mixer grinder.
  2. Water pump.
  3. Electric fan.
  4. Washing machine
  5. Vacuum cleaner.

28.) A bulb lights up when connected to a battery. State the energy change which takes place :

(i) in the battery.

(ii) in the bulb.

Answer: (i) When we plug on and bulb glow then Chemical energy in a battery changes into electrical energy.

(ii) When we plug on and bulb glow then electric energy is changes into light energy which produced by bulb.

29.) The hanging bob of a simple pendulum is displaced to one extreme position B and then released. It swings towards centre position A and then to the other extreme position C, In which position does the bob have :

(i) maximum potential energy ?

(ii) maximum kinetic energy ?

Give reasons for your answer.

Answer: When the bob is at mean position then total energy of the system is zero. But when we move the bob from mean position by applying external force then the potential energy of the bob increases with displacement from mean position.  If we released the bob then the potential energy decreases and kinetic energy increases as speed increases. This is illustrated as,

(i) Potential energy will be maximum at both extreme positions.

(ii) Kinetic energy will be maximum at mean position.

30.) A car of weight 20000 N climbs up a hill at a steady speed of 8 m/s, gaining a height of 120 m in 100 s. Calculate :

(a) work done by the car.

(b) power of engine of car.

Answer: Given, Weight or gravitational force on car = 20000 N. Speed 8 m/s,

Height or displacement=  120 m, time = 100 s.

If the car moves constant speed then net force on car must zero. So gravitational force on car is equal to upward force exerted on car by car.

Gravitational force = mechanical force exerted by car.

Total force = 20000 N.

(a) As we know,

Workdone = force × displacement.

Workdone = 20000 × 120

Workdone= 2400000

Workdone = 2400 × 10³ J.

Workdone = 2400 kJ.

The workdone by car is 2400 kJ.

(b) As we know,

Power = workdone / time.

Power = 2400 kJ/ 100 second.

Power= 24 kW

The power of engine of car is 24 kW.

Long Answer Type Questions:

31.) (a) What do you understand by the term “transformation of energy” ? Explain with an example.

(b) Explain the transformation of energy in the following cases :

(i) A ball thrown upwards.

(ii) A stone dropped from the roof of a building.

Answer: (a) According to law of conservation of energy, energy can transform into one form to another. In hydropower plant, gravitational potential energy is converted into electrical energy.  Then we use this electrical energy for fan. So fan can convert electrical energy into wind energy. Energy can only transformed into one form to another. So energy of the isolated system is constant.

(b) (i) A ball thrown upwards- When we throw a ball that time it had maximum kinetic energy and zero potential energy. As the ball increases the height there is increase in potential energy and decrease in kinetic energy. When it is at maximum height then it has maximum potential energy and zero kinetic energy.

(ii) A stone dropped from the roof of a building- when stone is at certain height then it possesses potential energy but value of kinetic energy is zero. But when it dropped then the value of speed increases because of gravitational force so value of kinetic energy increases and value of potential energy decreases. The body have maximum kinetic energy and zero potential energy at ground.

32.) (a) State and explain the law of conservation of energy with an example.

(b) Explain how, the total energy a swinging pendulum at any instant of time remains conserved. Illustrate your answer with the help of a labelled diagram.

Answer: (a)

Law of conservation of energy:

Statement- Law of conservation of energy states that energy neither created nor be destroyed but it can transform into one form to another. The total energy of an isolated system is constant.

Example: A body is thrown by applying certain external energy. When the ball moves upward then the value of potential energy increases and value of kinetic energy decreases but the value of total energy is constant.

(b) We can understand energy if swinging pendulum as follows.

Hang a pendulum with a string as shown in figure.

Apply external energy an moves the pendulum At a extreme position. At this point potential energy is maximum because potential energy is depend on displacement from mean position. When we release then potential energy goes on decreasing and kinetic energy increases because there is increase in speed. At mean position the value of kinetic energy becomes maximum and value of potential energy becomes zero.  Hence total energy remains constant.  We can plot energy curve for this.

33.) (a) What is the meaning of the symbol kWh ? What quantity does it represent ?

(b) How much electric energy in kWh is consumed by an electrical appliance of 1000 watts when it is switched on for 60 minutes ?

Answer: (a) The symbol kWh is kilowatt hour. It is the unit of energy.

As given,

Kilowatt hour = 10³ × watt × 1 hour.

As we know watt is the unit of power and hour is unit of time

Kilowatt hour = 10³ × power × time.

But power multiplied with time is the energy.

So kilowatt hour cam indicates the quantity of energy.

(b) Given, time 60 minutes = 1 hour, power = 1000 watt.

Energy = power × time

Energy = 1000 × 1

Energy = 1000 J

Energy = 1 KWh.

The total 1 kWh energy is consumed by that electric appliance.

34.) (a) Derive the relation between commercial unit of energy (kWh) and SI unit of energy (joule).

(b) A certain household consumes 650 units of electricity in a month. How much is this electricity in joules?

Answer: (a) Given, unit of energy is kWh.

We have to convert this into SI system. Kilo and hour are not SI units.

As we know,

1 kilo = 10³,

1 hour = 3600 seconds.

Put this value in Given unit.

1 kWh = 1 × 10³ × 3600 watt second.

Watt second is SI based unit.

1 kWh = 3.6 × 10⁶ watt second (J).

(b) Given, energy = 650 units.

1 unit = 1 kilowatt hour.

But we know that,

1 kWh = 3.6 × 10⁶ J.

So 1 unit = 3.6 × 10⁶ J.

Therefore 650 units =  650 × 3.6 × 10⁶ J

650 units = 2340 × 10⁶ J

650 units = 2.34 × 10⁹J.

That household consumes 2.34 × 10⁹ J of energy in a month.

35.) (a) Define power. Give the SI unit of power.

(b) A boy weighing 40 kg carries a box weighing 20 kg to the top of a building 15 m high in 25 seconds. Calculate the power. (g = 10 m/s²)

Answer: (a) Power: power is a rate of change of workdone with time. It gives information about rate of charge of energy. It is not depend on quantity of energy but it depends on rate of change of energy.

Power = energy/ time

Or

Power = workdone / time.

SI unit of power is J/ s means Watt.

(b) Given, Mass of the boy = 40 kg, Mass of box = 20 kg.

Total mass = 40 + 20 = 60 kg.

Height = 15 m, g = 10 m/s², time = 25 seconds

As we know that, when we do work against gravitational force then that workdone will stored in gravitational potential energy.

Gravitational potential energy = m × g × h

Gravitational potential energy = 60 × 10 × 15

Gravitational potential energy = 9000

As we know,

Power = energy / time

Power= 9000/ 25

Power = 360 watt

The value of power is 360 watt.

 

For more Chapter solution, click below


Updated: May 28, 2022 — 3:45 pm

2 Comments

Add a Comment
  1. Part 1 – que no 17 answer is wrong the Wright answer is 2J.

    So, Please do correction in que no 17

  2. Thanks to prepare my exam it was very easy to learn

Leave a Reply

Your email address will not be published.