Class 9 WORK

Topic covered

`color{red} ♦` INTRODUCTION
`color{red} ♦`WORK


In the previous few chapters we have talked about ways of describing the motion of objects, the cause of motion and gravitation.

Another concept that helps us understand and interpret many natural phenomena is ‘work’. Closely related to work are energy and power. In this chapter we shall study these concepts.

All living beings need food. Living beings have to perform several basic activities to survive. We call such activities ‘life processes’. The energy for these processes comes from food.

We need energy for other activities like playing, singing, reading, writing, thinking, jumping, cycling and running. Activities that are strenuous require more energy.

Animals too get engaged in activities. For example, they may jump and run. They have to fight, move away from enemies, find food or find a safe place to live. Also, we engage some animals to lift weights, carry loads, pull carts or plough fields.
All such activities require energy.

Think of machines. List the machines that you have come across. What do they need for their working? Why do some engines require fuel like petrol and diesel? Why do living beings and machines need energy?


There is a difference in the way we use the term ‘work’ in day-to-day life and the way we use it in science. To make this point clear let us consider a few examples.


Kamali is preparing for examinations. She spends lot of time in studies. She reads books, draws diagrams, organises her thoughts, collects question papers, attends classes, discusses problems with her friends, and performs experiments.

She expends a lot of energy on these activities. In common parlance, she is ‘working hard’. All this ‘hard work’ may involve very little ‘work’ if we go by the scientific definition of work.

You are working hard to push a huge rock. Let us say the rock does not move despite all the effort. You get completely exhausted. However, you have not done any work on the rock as there is no displacement of the rock.

You stand still for a few minutes with a heavy load on your head. You get tired. You have exerted yourself and have spent quite a bit of your energy. Are you doing work on the load?

The way we understand the term ‘work’ in science, work is not done.

You climb up the steps of a staircase and reach the second floor of a building just to see the landscape from there. You may even climb up a tall tree. If we apply the scientific definition, these activities involve a lot of work.

In day-to-day life, we consider any useful physical or mental labour as work. Activities like playing in a field, talking with friends, humming a tune, watching a movie, attending a function are sometimes not considered to be work.

What constitutes ‘work’ depends on the way we define it. We use and define the term work differently in science. To understand this let us do the following activities:

Activity _____________ `11.1`

♦ We have discussed in the above paragraphs a number of activities which we normally consider to be work in day-to-day life. For each of these activities, ask the following questions and answer them:

i) What is the work being done on?
(ii) What is happening to the object?
(iii) Who (what) is doing the work?


To understand the way we view work and define work from the point of view of science, let us consider some situations:

Push a pebble lying on a surface. The pebble moves through a distance. You exerted a force on the pebble and the pebble got displaced. In this situation work is done.

A girl pulls a trolley and the trolley moves through a distance. The girl has exerted a force on the trolley and it is displaced. Therefore, work is done.

Lift a book through a height. To do this you must apply a force. The book rises up. There is a force applied on the book and the book has moved. Hence, work is done.

A closer look at the above situations reveals that two conditions need to be satisfied for work to be done: (i) a force should act on an object, and (ii) the object must be displaced.

If any one of the above conditions does not exist, work is not done. This is the way we view work in science.

A bullock is pulling a cart. The cart moves. There is a force on the cart and the cart has moved. Do you think that work is done in this situation?

Activity _____________`11.2`

♦ Think of some situations from your daily life involving work
♦ List them.
♦ Discuss with your friends whether work is being done in each situation.
♦ Try to reason out your response.
♦ If work is done, which is the force acting on the object?
♦ What is the object on which the work is done?
♦ What happens to the object on which work is done?

Activity _____________`11.3`
♦ Think of situations when the object is not displaced in spite of a force acting on it.
♦ Also think of situations when an object gets displaced in the absence of a force acting on it.
♦ List all the situations that you can think of for each
♦ Discuss with your friends whether work is done in these situations.


How is work defined in science? To understand this, we shall first consider the case when the force is acting in the direction of displacement.

Let a constant force, F act on an object. Let the object be displaced through a distance, s in the direction of the force (Fig. 11.1). Let W be the work done. We define work to be equal to the product of the force and displacement.

Work done = force × displacement

`W = F s` ..........(11.1)

Thus, work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction.

In Eq. (11.1), if `F = 1 N` and `s = 1 m` then the work done by the force will be `1 N m`. Here the unit of work is newton metre `(N m)` or joule (J). Thus `1 J` is the amount of work done on an object when a force of `1 N` displaces it by `1 m` along the line of action of the force.

Look at Eq. (11.1) carefully. What is the work done when the force on the object is zero? What would be the work done when the displacement of the object is zero? Refer to the conditions that are to be satisfied to say that work is done.

Q 3255201164

A porter lifts a luggage of `15 kg` from the ground and puts it on his head `1.5 m` above the ground. Calculate the work done by him on the luggage.
Class 9 Chapter 11 Example 1

Mass of luggage, `m = 15 kg` and
displacement, `s = 1.5 m`.

Work done, `W = F xx s = mg xx s`
`= 15 kg xx 10 m s^(-2) xx 1.5 m`
`= 225 kg m s^(-2) m`
`= 225 N m = 225 J`
Work done is `225 J`.

another situation

Consider another situation in which the force and the displacement are in the same direction: a baby pulling a toy car parallel to the ground, as shown in Fig. 11.4. The baby has exerted a force in the direction of displacement of the car.

In this situation, the work done will be equal to the product of the force and displacement. In such situations, the work done by the force is taken as positive.

Consider a situation in which an object is being displaced by the action of forces and we identify one of the forces, F acting opposite to the direction of the displacement s, that is, the angle between the two directions is 180º.

In such a situation, the work done by the force, F is taken as negative and denoted by the minus sign. The work done by the force is `F xx (–s)` or `(–F xx s).`

It is clear from the above discussion that the work done by a force can be either positive or negative. To understand this, let us do the following activity:

Activity _____________`11.4`

♦ Lift an object up. Work is done by the force exerted by you on the object. The object moves upwards. The force you exerted is in the direction of displacement. However, there is the force of gravity acting on the object.
♦ Which one of these forces is doing positive work?
♦ Which one is doing negative work?
♦ Give reasons.

Work done is negative when the force acts opposite to the direction of displacement. Work done is positive when the force is in the direction of displacement.
Q 3265201165

An object of mass `15 kg` is moving with a uniform velocity of `4 m s^(–1)`. What is the kinetic energy possessed by the object?

Class 9 Chapter 11 Example 3

Mass of the object, `m = 15 kg`, velocity
of the object, `v = 4 m s^(–1)`.
From Eq. (11.5),

` E_k = 1/ 2 mv^2`

` = 1/2 xx 15 kg xx 4 m s^(–1) xx 4 m s^(–1)`

`= 120 J`
The kinetic energy of the object is `120 J`.