## Work and Energy Class 9 Science Notes Maharashtra State Board

Generally, any mental or physical activity is referred to as work. When we walk or run, the energy in our body is used to do the necessary work. We say that a girl who is studying is working or performing work. But that is mental work. In physics, we deal with physical work. Work has a special meaning in physics.

‘Work is said to be done when a force applied on an object causes displacement of the object.’

You have already learned that the work done by a force acting on an object is the product of the magnitude of the force and the displacement of the object in the direction of the force. Thus, Work = force × displacement

Minakshee wants to displace a wooden block from point A to point B along the surface of a table as shown in the figure. She has used force F for the purpose. Has all the energy she spent been used to produce acceleration in the block? Which forces have been overcome using that energy?

You must have seen the events depicted in pictures B and C above. When a child pulls a toy with the help of a string, the direction of the force is different from that of displacement. Similarly, when a large vehicle tows a small one, the directions of force and the displacements are different. In both these cases, the direction of force makes an angle with the direction of displacement. Let us see how to calculate work done in such cases.

When a child pulls a toy cart, force is applied along the direction of the string while the cart is pulled along the horizontal surface. In this case, to calculate the amount of work done, we have to convert the applied force into the force acting along the direction of displacement.

Let F be the applied force and F_{1} be its component in the direction of displacement. Let s be the displacement. The amount of work done is given by

W = F_{1}.s …………..(1)

The force F is applied in the direction of the string i.e. at an angle with the horizontal. Let q be the angle that the string makes with the horizontal. We can determine the component F_{1}, of this force F, which acts in the horizontal direction using trigonometry (see figure).

cos θ = base/hypotenuse

cos θ = F_{1}/F

F_{1} = F cos θ

Thus, the work done by F_{1} is

W = F cos θ s

W = F s cos θ

**Unit of Work**

Work = Force × Displacement

In the SI system, the unit of force is Newton (N) and the unit of displacement is meter (m). Thus, the unit of force is Newton-metre. This is called joule.

1 Joule: If a force of 1 newton displaces an object through 1 meter in the direction of the force, the amount of work done on the object is 1 joule.

∴ 1 joule = 1 newton × 1 metre

1 J = 1 N × 1 m

In the CGS system, the unit of force is dyne and that of displacement is centimeter (cm). Thus, the unit of work done is dyne-centimetre. This is called an erg.

1erg: If a force of 1 dyne displaces an object through 1 centimeter in the direction of the force, the amount of work done is 1 erg.

1 erg = 1 dyne × 1 cm

**Relationship between Joule and Erg**

We know that, 1 newton = 10^{5} dyne and 1 m = 10^{2} cm

Work = force × displacement

1 joule = 1 newton × 1 m

1 joule = 10^{5} dyne × 10^{2} cm = 10^{7} dyne cm

1 joule = 10^{7} erg

**Positive, Negative, and Zero Work**

- Pushing a stalled vehicle
- Catching the ball that your friend has thrown towards you.
- Tying a stone to one end of a string and swinging it round and round by the other end of the string.
- Walking up and down a staircase; climbing a tree.
- Stopping a moving car by applying brakes.

You will notice that in some of the above examples, the direction of the force and displacement are the same. In some other cases, these directions are opposite to each other, while in some cases, they are perpendicular to each other. In these cases, the work done by the force is as follows.

- When the force and the displacement are in the same direction (θ = 0°), the work done by the force is positive.
- When the force and the displacement are in opposite directions (θ = 180°), the work done by the force is negative.
- When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other (θ = 90°), the work done by the force is zero.

Take a plastic cup and make a hole in the center of its bottom. Take a long thread, double it, and pass it through the hole. Tie a thick enough knot at the end so that the knot will not pass through the hole, taking care that the two loose ends are below the bottom of the cup. Tie a nut each to the two ends as shown in the figure. Now do the following.

- As shown in Figure ‘A’, put the cup on a table, keep one of the nuts in the cup, and let the thread carrying the other nut hang down along the side of the table. What happens?
- As shown in Figure ‘B’, when the cup is sliding along the table, stop it by putting a ruler in its path.
- As shown in Figure ‘C’, keep the cup at the center of the table and leave the two nuts hanging on opposite sides of the table.

Suppose an artificial satellite is moving around the earth in a circular orbit. As the gravitational force acting on the satellite (along the radius of the circle) and its displacement (along the tangent to the circle) are perpendicular to each other, the work done by the gravitational force is zero.

**Institutes at Work**

The National Physical Laboratory, New Delhi, was conceptualized in 1943. It functions under the Council of Scientific and Industrial Research. It conducts basic research in the various branches of physics and helps various industries and institutes engage in developmental work. Its main objective is to establish national standards for various physical quantities.

**Energy**

Why does it happen?

- If a pot having a plant is kept in the dark, the plant languishes.
- On increasing the volume of a music system or TV beyond a limit, the vessels in the house start vibrating.
- Collecting sunlight on a paper with the help of a convex lens burns the paper.

The capacity of a body to perform work is called its energy. The units of work and energy are the same. The unit in the SI system is joule while that in the CGS system is erg.

You have learned that energy exists in various forms like mechanical, heat, light, sound, electromagnetic, chemical, nuclear, and solar. In this chapter, we are going to study two forms of mechanical energy, namely, potential energy and kinetic energy.

**Kinetic Energy**

What will happen in the following cases?

- A fast cricket ball strikes the stumps.
- The striker hits a coin on the carom board.
- One marble strikes another in a game of marbles.

From the above examples we understand that when a moving object strikes a stationary object, the stationary object moves. Thus, the moving object has some energy, part or all of which it shares with the stationary object, thereby setting it in motion. ‘The energy which an object has because of its motion is called its kinetic energy’. The work done by a force to displace a stationary object through a distance s is the kinetic energy gained by the object.

Kinetic energy = work done on the object

∴ K.E. = F × s

**Expression for Kinetic Energy:**

Suppose a stationary object of mass m moves because of an applied force. Let u be its initial velocity (here u = 0). Let the applied force be F. This generates an acceleration ‘a’ in the object, and, after time t, the velocity of the object becomes equal to v. The displacement during this time is s. The work done on the object, W = F.s

W = F × s

According to Newton’s second law of motion,

F = ma ……….(1)

Similarly, using Newton’s second equation of motion

s = ut + \(\frac{1}{2}\)at^{2}

However, as initial velocity is zero, u = 0.

s = 0 + \(\frac{1}{2}\)at^{2}

s = \(\frac{1}{2}\)at^{2} ………..(2)

∴ W = ma × \(\frac{1}{2}\)at^{2} (using equations (1) and (2))

W = \(\frac{1}{2}\)m(at)^{2} ……….(3)

Using Newton’s first equation of motion

v = u + at

∴ v = 0 + at

∴ v = at

∴ v^{2} = (at)^{2} ……..(4)

∴ W = \(\frac{1}{2}\)mv2 (using equations (3) and (4))

The kinetic energy gained by an object is the amount of work done on the object.

∴ K.E = W

∴ K.E = \(\frac{1}{2}\)mv^{2}

**Potential Energy**

‘The energy stored in an object because of its specific state or position is called its potential energy.’

- Hold a chalk at a height of 5 cm from the floor and release it.
- Now stand up straight and then release the chalk.
- Is there a difference in the results of the two activities? If so, why?

**Expression for Potential Energy**

To carry an object of mass ‘m’ to a height ‘h’ above the earth’s surface, a force equal to ‘mg’ has to be used against the direction of the gravitational force. The amount of work done can be calculated as follows.

Work = force × displacement

W = mg × h

∴ W = mgh

∴ The amount of potential energy stored in the object because of its displacement

P.E = mgh (W = P.E)

∴ Displacement to height h causes energy equal to mgh to be stored in the object.

Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this?

Potential energy is relative. The heights of the ball concerning Ajay and Atul are different. So the potential energy concerning them will be different.

**Transformation of Energy**

Energy can be transformed from one type to another. For example, exploding firecrackers convert the chemical energy stored in them into light, sound, and heat energy.

Observe the above diagram and discuss how the transformation of energy takes place, giving examples of each.

**Law of Conservation of Energy**

‘Energy can neither be created nor destroyed. It can be converted from one form into another. Thus, the total amount of energy in the universe remains constant.

Make two pendulums of the same length with the help of thread and two nuts. Tie another thread in the horizontal position. Tie the two pendulums to this horizontal thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and observe. What do you see?

You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other.

**Free Fall**

If we release an object from a height, it gets pulled towards the earth because of the gravitational force. An object falling solely under the influence of gravitational force is said to be in free fall or to be falling freely. Let us look at the kinetic and potential energies of an object of mass m, falling freely from height h when the object is at different heights.

As shown in the figure, the point A is at a height h from the ground. Let the point B be at a distance x, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B, and C.

1. When the object is stationary at A, its initial velocity is u = 0

∴ K.E. = mass × velocity^{2} = \(\frac{1}{2}\)mu^{2}

K.E. = 0

P.E. = mgh

∴ Total energy = K.E. + P.E. = 0 + mgh

Total Energy = mgh ……….(1)

2. Let the velocity of the object be v_{B} when it reaches point B, having fallen through a distance x.

u = 0, s = x, a = g

v^{2} = u^{2} + 2as

\(\mathrm{V}_{\mathrm{B}}^2\) = 0 + 2gx

\(\mathrm{V}_{\mathrm{B}}^2\) = 2gx

∴ K.E. = \(\frac{1}{2}{m v_B}^2\) = \(\frac{1}{2}\)m(2gx)

K.E. = mgx

Height of the object when at B = h – x

∴ P.E. = mg(h – x)

∴ P.E. = mgh – mgx

∴ Total Energy T.E. = K.E. + P.E. = mgx + mgh – mgx

∴ T.E. = mgh ………(2)

3. Let the velocity of the object be v_{C} when it reaches the ground, near point C.

u = 0, s = h, a = g

v^{2} = u^{2} + 2as

\(\mathrm{V}_{\mathrm{C}}^2\)= 0 + 2gh

∴ K.E. = \(\frac{1}{2}{m v_C}^2\) = \(\frac{1}{2}\)m(2gh)

K.E. = mgh

The height of the object from the ground at point C is h = 0

∴ P.E. = mgh = 0

∴ T.E. = K.E. + P.E

∴ T.E. = mgh ……….(3)

From equations (1), (2), and (3) we see that the total energy of the object is the same at the three points A, B, and C. Thus, every object has potential energy when it is at a height above the ground and it keeps getting converted to kinetic energy as the object falls towards the ground. On reaching the ground (point C), all the potential energy gets converted to kinetic energy. But at any point during the fall the total energy remains constant.

i.e., T.E. = P.E. + K.E.

T.E. at A = mgh + 0 = mgh

T.E. at B = mgx + mg (h – x) = mgh

T.E. at C = 0 + mgh = mgh

**Power**

- Can your father climb stairs as fast as you can?
- Will you fill the overhead water tank with the help of a bucket or an electric motor?
- Suppose Rajashree, Yash, and Ranjeet have to reach the top of a small hill. Rajashree went by car, Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last?

In the above examples, the work done is the same in each example but the time taken to perform the work is different for each person or each method. The fast or slow rate of the work done is expressed in terms of power. ‘Power is the rate at which work is done.’

If W amount of work is done in time t then,

Power = \(\frac{\text { Work }}{\text { Time }}\)

P = \(\frac{W}{t}\)

In the SI system, the unit of work is J, so the unit of power is J/s. This is called watt

1 watt = 1 joule/1 second

In the industrial sector, the unit used to measure power is called ‘horsepower.’

1 horse power = 746 watt.

The unit of energy for commercial use is kilowatt hour.

1000 joule work performed in 1 second is 1 kilowatt power.

1 kW h = 1 kW × 1hr

= 1000 W × 3600 s

= 3600000 J

1 kWh = 3.6 × 10^{6} J

Electricity used for domestic purposes is measured in units of kilowatt hour.

1 kWh = 1 unit

Good Maharashtra State Board Class 9 Science Notes Work and Energy can simplify complex concepts and make studying more efficient.