# Chapter 5 – Work Power and Energy | Chapter Solution Class 9

 Publisher : Santra publication pvt. ltd. Book Name : Madhyamik Physical Science And Environment Class : 9 (Madhyamik) Subject : Physical Science Chapter Name : Work Power and Energy

## Multiple Choice Questions (MCQ)

[Each of Mark-1]

Question 1

The rate of doing work is called

1. velocity
2. energy
3. power
4. force

power

Explanation

The rate of doing work is called power, which measures how quickly energy is transferred or work is performed.

Question 2

The unit of energy in the SI system is

1. erg
2. watt
3. horsepower
4. joule

joule

Explanation

The unit of energy in the SI system is the joule, which represents the amount of energy transferred or work done by the force of one Newton over a distance of one meter.

Question 3

One watt is equal to

1. 1 erg/s
2. 1 g.cm/s
3. 1 kg-m/s
4. 1 joule/s

1 joule per second (J/s)

Explanation

One watt is equal to 1 joule per second (J/s), representing the rate at which energy is transferred or work is done.

Question 4

One horsepower is equal to

1. 500 W
2. 1000 W
3. 746 W
4. 764 W

746 watts (W)

Explanation

One horsepower is equal to 746 watts (W), which is a unit of power used to measure the rate of work done or energy transferred.

Question 5

Kilowatt is the unit of

1. energy
2. work done
3. Power
4. force

power

Explanation

Kilowatt (kW) is the unit of power, which measures the rate at which work is done or energy is transferred.

Question 6

In the case of negative work, the angle between the force and displacement is

1. 45°
2. 90°
3. 180°

180°

Explanation

In the case of negative work, the angle between the force and displacement is 180°, indicating they act in opposite directions.

Question 7

In a tug of war, work done by the losing team

1. positive
2. negative
3. zero
4. none of these

zero

Explanation

In a tug of war, the work done by the losing team is zero, as there is no displacement in their direction of force.

Question 8

The work done on an object does not depend upon the

1. displacement
2. the initial velocity of the object
3. force applied
4. the angle between the force and the displacement

the initial velocity of the object

Explanation

The work done on an object does not depend upon the initial velocity of the object. Work depends on displacement, force applied, and the angle between force and displacement.

Question 9

Which one of the following is not the unit of energy?

1. kilowatt
2. joule
3. newton-metre
4. kilowatt-hour

kilowatt-hours

Explanation

Kilowatt (kW) is not a unit of energy; it is a unit of power. Energy is measured in joules (J) or kilowatt-hours (kWh).

Question 10

When an object falls freely under gravity, then its total energy

1. increases
2. decreases
3. remains constant
4. first increases and then decreases

remains constant

Explanation

When an object falls freely under gravity, its total energy remains constant as potential energy decreases, while kinetic energy increases correspondingly, maintaining the total energy.

## Fill in the blanks

[Each of Mark-1]

1. The rate of doing work is called ____.
2. The SI unit of work is Joule ____.
3. Kilowatt-hour is the unit of ____.
4. 1 watt is a rate of working of one per ____ per ____
5. The capacity of doing work is called ____
6. Power = force × ____.
7. Due to a change in configuration, a body acquires ____ energy.
8. Both power and work are ____ quantities.
9. Dimension of energy is of ____.
10. In case of a freely falling body, total mechanical energy is always ____.

1. Power
2. Joule (J)
3. Energy
4. second
5. Energy
6. velocity or speed
7. potential energy
8. scalar quantities
9. M L2 T-2
10. conserved

## Answer in one word or in one sentence

[Each of Mark-1]

1. What is the SI unit of work?
2. Cite an example of no workforce.
3. What is the relation between joule and erg?
4. Name one scalar quantity whose unit contains three fundamental units.
5. Write down the SI unit of kinetic energy.
6. How many watts equal one horsepower?
7. Name two units of power bigger than a watt.
8. What is the work done by the force of gravity on a satellite moving around the earth?
9. Identify the kind of mechanical energy possessed by a running athlete.
10. Name the commercial unit of energy.

1. Joule.
2. Friction.
3. 1 Joule = 107 ergs.
4. Speed.
5. Joule (J).
6. Approximately 745.7 watts.
7. Kilowatt (kW) and Megawatt (MW).
8. Zero (no work is done by gravity on a satellite moving in a circular orbit).
9. Kinetic energy.
10. Kilowatt-hour (kWh).

[Each of Mark-2]

Question 1

Define work. How is work done measured ?

Work is the transfer of energy that occurs when a force is applied to an object, causing displacement in the direction of the force. Work done is measured as the product of the applied force and the distance over which the force is applied.

Question 2

Write down the dimension of the work. Is work a scalar or a vector quantity?

The dimension of work is [ML2T-2]. Work is a scalar quantity.

Question 3

Under what conditions work is not done although force is acting?

Work is not done when the force acting on an object is perpendicular to the direction of displacement.

Question 4

Define no work force with an example.

A no work force is a force that does not cause any displacement, resulting in zero work. For example, pushing against a stationary wall with all your strength does not cause any displacement and therefore no work is done.

Question 5

Establish the relation between (a) joule and erg and (b) kg-m and joule.

(a) Relation between joule (J) and erg (erg):

1 Joule = 1 N-m = 1 (105 dyne) × (102 cm)

1 Joule = 1 (105 × 102) dyne-cm

1 Joule = 107 ergs

Therefore, 1 Joule is equal to 107 ergs.

(b) Relation between kilogram-meter (kg-m) and joule (J):

Work = Force × Distance

Force = Mass × Acceleration

Therefore,

Work = Mass × Acceleration × Distance

For the kg-m unit of work, the mass is 1 kilogram, the acceleration is 9.81 m/s^2, and the distance is 1 meter. Therefore, the work done is equal to 9.81 joules.

1 kg-m = 9.81 J

Question 6

Define power. What is its relation with work?

Power is the rate at which work is done or energy is transferred. It measures how quickly work is performed.

In equation form:

Power = Work / Time

Question 7

Establish the relation between power and force.

Power is defined as the rate at which work is done:

Power = Work \over Time

P = F × d \over t   [W = F × d]

P = F × v

Therefore, the relation between power (P), force (F), and velocity (v) is given by:

Power = Force × Velocity

Question 8

Define mechanical energy. Classify it.

Mechanical energy refers to the sum of potential energy and kinetic energy possessed by an object due to its motion or position. It is associated with the movement or configuration of objects and is a form of energy that can be converted into other forms.

Mechanical energy can be classified into two types:

1. Kinetic Energy: Kinetic energy is the energy possessed by an object due to its motion.
2. Potential Energy: Potential energy is the energy possessed by an object due to its position or configuration.

Question 9

Define kinetic energy and potential energy.

1. Kinetic Energy: Kinetic energy is the energy possessed by an object due to its motion.
2. Potential Energy: Potential energy is the energy possessed by an object due to its position or configuration.

Question 10

Explain whether work is done due to the revolution of the earth around the sun.

No, work is not done due to the revolution of the Earth around the Sun.

Work is a measure of the energy transferred when a force causes an object to move in the direction of that force. When the Earth revolves around the Sun, there is a gravitational force between them, but this force acts perpendicular to the direction of motion. [W = FS cos 90º = 0]

Question 11

What change of kinetic energy of a body will be observed when (i) its mass is doubled, (ii) its velocity is doubled?

1. When the mass of a body is doubled, the kinetic energy doubles if the velocity remains constant.
2. When the velocity of a body is doubled, the kinetic energy increases by a factor of four if the mass remains constant.

Question 12

What do you mean by gravitational potential energy? On what factors does it depend?

Gravitational potential energy refers to the energy possessed by an object due to its position in a gravitational field. It arises from the gravitational attraction between the object and the Earth (or any other massive object).

Gravitational potential energy depends on two factors:

1. Height or elevation
2. Mass

[Each of Mark-3]

Question 1

Define work. What do you mean by work done by a force and work done against a force ? Give examples of each.

Work is the energy transferred when a force causes an object to move.

Work done by a force: When a force applied to an object causes it to move in the same direction as the force, work is done by the force. Example: Pushing a car forward.

Work done against a force: When a force opposes the motion of an object, work is done against the force. Example: Lifting a heavy weight against gravity.

Question 2

Define the term ‘energy’. Write the names of different forms of energy. Establish an expression for gravitational potential energy.

Energy is the ability or capacity to do work.

Different forms of energy include:

1. Kinetic Energy: Energy associated with the motion of an object.
2. Potential Energy: Energy stored in an object based on its position or state.

Gravitational potential energy is the energy possessed by an object due to its position above the ground. The expression for gravitational potential energy (PE) is:

PE = m × g × h

Question 3

Define no work force with an example. State two factors on which magnitude of work depends.

A “no work force” refers to a force that does not cause any displacement of an object when applied. This means that the force does not transfer any energy or do any work on the object.

Example: Consider a person pushing against a wall. Despite applying force, the wall does not move, and therefore, no work is done on the wall. The force exerted by the person does not result in any displacement, so no work is performed.

The magnitude of work depends on two factors:

1. The magnitude of the force: The greater the magnitude of the force applied to an object, the more work is done.
2. Displacement of the object: Work is directly proportional to the distance the object moves in the direction of the force. The larger the displacement, the more work is done.

Question 4

Define kinetic energy. Derive an expression for the kinetic energy of an object of mass m in moving with a velocity v.

Kinetic energy is the energy possessed by an object due to its motion.

The expression for the kinetic energy (KE) of an object of mass m moving with a velocity v is:

KE = {1\over2} × m × v^2

Derivation

Kinetic energy (Ek) = work done

= Force × displacement

= F×s

= m × a × s

= m × v^2\over 2 [∵ v2 = u2 + 2as and in case initial velocity (u) = 0 ∴ v2 = 2as or as v^2\over 2]

∴ KE = {1\over2} × m × v^2

∴ Kinetic Energy = {1\over2} × mass × velocity^2

Question 5

State the law of conservation of mechanical energy. Show that the total mechanical energy of a freely falling body is conserved.

In a conservative force system, the total mechanical energy i.e. the sum of kinetic and potential energy is always constant – this is known as the law of conservation of mechanical energy.

The above law is valid for a freely falling body under gravity.

Let a body of mass m is raised to a vertical height h above the ground and it is kept rest at A. At point A, the energy of the body is entirely potential and it is EA = mgh.

Now, when the body falls vertically downloads, its velocity goes on increasing and when the body is at point B (when it falls a vertical height x from A) let its velocity become VB. From the relation v2 = u2 + 2gh

we get, vB2 = 02 + 2gx  ∴ vB2 = 2gx

At point B, the energy of the body is partly potential and partly kinetic.

∴ Total energy at B is EB = mg(h-x) + 1\over 2mvB2

= mg(h-x) + 1\over 2m2gx

= mgh – mgx + mgx

= mgh

Again, when the body reaches point C (just before hitting the ground) it possesses kinetic energy only and if the velocity of the body at that point C be then v then vc2 = 2gh

∴ Total Energy at C, Ec = 1\over 2 mv2

= 1\over 2m2gh

= mgh

∴ EA = EB = EC = mgh

i.e. total mechanical energy at all points during the free fall of a body under gravity remains constant. Hence the principle of conservation of energy is valid for a freely falling body.

## Numerical problems

Question 1

Applying a force of 20 N, a body is moved through 10 m in the direction of the force. Find the amount of work done. [Ans. 200 J]

Force (F) = 20 N

Displacement (s) = 10 m

Work = F × s = 20 × 10 = 200 J

Question 2

How many joules of work is performed by a machine of power 40 W in 1{1\over2} hour? [Ans. 216000 J]

Power (P) = 40 W

Time (t) = 1{1\over2} hour = 1.5 × 60 × 60 = 5400 s

Work (W) = Power (P) × Time (t)

= 40 × 5400 = 216000 J

Question 3

A man of mass 50 kg ascends 30 steps of a staircase in 15 seconds. If the height of each step of the staircase is 20 cm, find the power of the man (g = 9.8 m/s2). [Ans. 196 W]

Mass (m) = 50 kg

Height (h) = 30 × 20 cm = 6 m

Time (t) = 15 s

Step 1: Calculation of work

Work (W) = mgh

= 50 × 9.8 × 6 = 2940 J

Step 2: Calculation of Power

Power (P) = Work\over Time

= 2940\over 15 = 196 W

Question 4

A car of weight 20000 N drives up a hill at a steady speed of 8 m/s, gaining a height of 120 m in 100 s. Calculate (i) the work done by the car and (ii) the power of the engine of the car. [(i) 24 × 105 J (ii) 24 × 103 W]

Force (F) = 20000 N

Speed (v) = 8 m/s

Height (h) = 120 m

Time (t) = 100 s

(i) Work done (W) = Force (F) × displacement (h)

= 20000 × 120

= 2400000 J or 2.4 × 106 J

(ii) Power (P) = Work\over Time

= 2400000\over 100

= 24000 W or 2.4 × 104 W

Question 5

A force of 10 N-acting on a body at an angle of 30° with the horizontal direction displaces it horizontally through a distance of 6 m. Calculate the work done. [Ans. 30 √3 J]

Force (F) = 10 N

Displacement (s) = 6 m

Angle (θ) = 30°

Work (W) = Force (F) × Displacement (s) cos θ

= 10 × 6 × cos 30°

= 60 × √3 \over 2

= 30 √3 J

Question 6

A bag of wheat weighs 75 kg. Calculate the height to which it should be raised so that its potential energy is 7350 J. (Given g = 9.8 m/s2) [Ans. 10 m]

Mass (m) = 75 kg

Height (h) = To be calculated

Potential Energy (PE) = 7350 J

Now, PE = mgh

height (h) = PE\over mg

= 7350\over 75 × 9.8

= 10 m

Question 7

A moving body of mass 10 kg has 20 Joules of kinetic energy. Calculate the speed of the body. [Ans. 2 m/s]

Mass (m) = 10 kg

Kinetic Energy (KE) = 20 J

Speed (v) = To be calculated

Kinetic Energy (KE) = {1\over2}×m×v^2

or, {1\over2}×10×v^2 = 20

or, v2 = 20 × 2\over 10

or, speed (v) = √4 = 2 m/s

Question 8

How much water per minute a pump of power 2 kW can raise to a height of 10 m? (g = 10 m/s2) [Ans. 1200 kg]

Power (P) = 2 kW = 2000 W

Time (t) = 1 min =  60 s

Height (h) = 10 m

Step 1: Calculation of Work

Work (W) = m×g×h

= 10 × 10 × m

= 100 m

Power (P) = Work\over time

or, 2000 =  100 m\over 60

or, 100 m\over 60 = 2000

or, m = 2000 × 60\over 100

or, mass (m) = 1200 Kg

Question 9

How much kilogram weight will a man working at the power of 100 W be able to rise at a constant speed of 1 m/s vertically? (Given g = 10 m/s2) [Ans. 10 kg]

Power (P) = 100 W

Speed (v) = 1 m/s

Step 1: Calculation of Force

Power (P) = Force (F) × Speed (v)

or, Force (F) = Power\over speed

or, Force (F) = 100\over 1

or, Force (F) = 100 N

Step 2: Calculation of Mass

Force (F) = mass (m) × acceleration due to gravity (g)

or, mass (m) = Force\over g

or, mass (m) = 100\over 10

or, mass (m) = 10 kg

Question 10

Calculate the potential energy of a body of mass 50 kg raised to a height of 4 m above the ground. If the body falls down under gravity freely, then what will be its velocity when it just touches the ground? (Given g = 9.8 m/s2) [Ans. 1960 J, 8.9 m/s (approx)]

Mass (m) = 50 kg

Height (h) = 4 m and g = 9.8 m/s2

(i) Calculation of Potential Energy

PE = m×g×h

= 50×9.8×4

= 1960 J

(ii) Calculation of velocity

Kinetic Energy = Potential Energy

or, {1\over2}×m×v^2 = 1960

or, {1\over2}×50×v^2 = 1960

or, v2 = 1960×2\over 50

or, v = √78.4 = 8.85 m/s

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