CBSE Class 9 Science Work and Energy Exam Notes

CBSE Class 9 Science Work and Energy Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

WORK

Definition : In our daily life "work" implies an activity resulting in muscular or mental exertion. However, in physics the term 'work' is used in a specific sense involves the displacement of a particle or body under the action of a force. "work is said to be done when the point of application of a force moves.

 Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force.

Work = Force × Distance moved in the direction  of force.

The work done by a force on a body depends on two factors :

 (i) Magnitude of the force, and

(ii) Distance through which the body moves (in the direction of force)

Unit of Work

When a force of 1 newton moves a body through a distance of 1 meter in its own direction, then the work done is known as 1 joule.

Ex.1How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?

Sol The work done is calculated by using the formula:

                                                                      W = F × S

Here, Force, F = 10 N And, Distance, S = 1 m

So, Work done, W = 10 × 1 J

= 10 J

Thus, the work done is 10 joules

Ex.2 Find the work done by a force of 10 N in moving an object through a distance of 2 m.

Sol. Work done = Force × Distance moved Here, Force = 10 N Distance moved = 2 m

Work done, W = 10 N × 2 m = 20 Joule = 20 J

WORK DONE ANALYSIS

♦ Work done when force and displacement are along same line.

♦ Work done by a force : Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.

♦ Work done against the force : Work is said to be done against a force if the direction of the displacement is opposite to that of the force.

 Work done against Gravity : To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If 'm' is the mass of the object and 'h' is the height through which it is raised, then the upward force

(F) = force of gravity = mg

If 'W' stands for work done, then

W = F . h = mg . h

Thus W = mgh

Therefore we can say that, "The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted.

Ex.3 Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.

Sol. Force, F = 120 N; Distance, s = 100 m

Using the formula, we have

W = Fs = 120 N × 100 m = 12,000 J

Ex.4 A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s2. Calculate the work done.

Sol. Given :mass, m = 5 kg acceleration, a = 3 m/s2

Force acting on the body is given by

F = ma = 5 × 3 = 15 N Now, work done is given by

W= Fs = 15 N × 4 m = 60 J

Ex.5 Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms^–2).

Sol. Force of gravity

mg = 200 × 9.8 = 1960.0 N

h = 5 m

Work done, W = mgh

or W= 1960 × 5 = 9800 J

♦ Work done when force and displacement are inclined (Oblique case)

Consider a force 'F' acting at angle q to the direction of displacement 's' as shown in fig.

♦ Work done when force is perpendicular to Displacement

q = 90º

W = F.S × cos 90º = F.S × 0 = 0

Thus no work is done when a force acts at right angle to the displacement.

♦ Special Examples :

 When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.

 Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.

 A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.

Ex.6 A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60º with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done.

Sol. Given F = 100 N, s = 3 m, q = 60º.

Work done is given by

W = Fs cos q = 100 × 2 × cos 60º

= 100 × 3 × 1/2

1 = 150 J (Q cos 60º = 1/2 )

Ex.7 An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.

Ø POWER

♦ Definition : Power is defined as the rate of doing work

Power = Work done /T ime taken

 P = W/t

In other words, power is the work done per unit time, power is a scalar quantity.

Since W = F.S therefore

P = =FS/t = F × V = force × velocity

Unit of power : The S.I. unit of power is watt and it is the rate of doing work at 1 joule per second.

1 watt =

1 joule

1seconds

1 kilowatt = 1 kW = 1000 W

1 Horse power = 1 H.P. = 746 W

Ex.8 A machine raises a load of 750 N through a height of 15 m in 5s. Calculate : (i) the work done by the machine.

(ii) the power at which the machine works.

Sol. (i) Work done is given by W = F.s

Here F = 750 N; s = 15 m

 W = 750 × 15 = 11250 J

= 11.250 kJ

(ii) Now, power of the machine is given by

P =W/t

Here, W = 11250 J; t = 5 s

Power P = 11250J /5s = 2250 W = 2.250 kW

Ex.9 A weight lifter lifted a load of 100 kg to a height of 3 m in 10 s. Calculate the following: (i) amount of work done

(ii) power developed by him

Sol. (i) Work done is given by

W = F . s

Here, F = mg = 100 × 10 = 1000 N

W = 1000 N × 3 m = 3000 joule

(ii) Now, P = W/t , where W = 3000 J and t = 10 s

 P = 3000 J/10 s = 300 W

Ex.10 A water pump raises 60 liters of water through a height of 20 m in 5 s. Calculate the power of the pump. (Given: g = 10 m/s2, density of water = 1000 kg/m³)

Sol. Work done, W = F.s ...(1)

Here, F = mg ...(2) But, Mass = volume × density

Volume = 60 liters = 60 × 10–3 m3

Density = 1000 kg/m3

 Mass , m = (60 × 10–3 m3) × (1000 kg/m3) = 60 kg

 Equation (2) becomes

F = 60 kg × 10 m/s² = 600 N

Now, W = F . s = 600 N × 20 m = 12000 J \ Power = W/t = 12000J/5s= 2400 W