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Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.

## Units of Energy and Formula for Kinetic Energy and Potential Energy

If a person can do a lot of work we say that he has a lot of energy or he is very energetic. In physics also, anything which is able to do work is said to possess energy. Thus, energy is the ability to do work. Let us take one example to understand it more clearly. To cut a log of wood into small pieces, we have to raise the axe vertically above the log of wood and some work has to be done in raising the axe.

If the axe is now allowed to fall on wood, it can do work in cutting the wood. Thus, the work done in raising the axe has been stored up in it, giving it the ability for doing work. Now, when the axe is resting on the log of wood, it can no longer do any work. To give it the ability to do work again, work has to be done in raising it above the log of wood once again.

We say that the raised axe has the energy or ability for doing work. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. It should be noted that whenever work is done, energy is consumed.

A body having energy can do work as follows : A body which possesses energy can exert a force on another object, body is transferred to the object. By gaining energy, the object moves. And when the object moves, work is said to be done. Energy is a scalar quantity. It has only magnitude but no direction.

### Unit of Energy

The units of work and energy are the same. So, the SI unit of energy is joule (which is denoted by the letter J). Whenever work is done, an equal amount of energy is consumed. Keeping this is mind, we can define 1 joule energy as follows :

The energy required to do 1 joule of work is called 1 joule energy. Joule is a small unit of energy, so sometimes a bigger unit of energy called ‘kilojoule’ is also used. The symbol of kilojoule is kJ. Now,

1 kilojoule = 1000 joules

or 1 kJ = 1000 J

The unit of energy called ‘joule’ is named after a British physicist James Prescott Joule.

### Different Forms of Energy

Energy exists in many forms. The main forms of energy are :

- Kinetic energy
- Potential energy
- Chemical energy nv:studygear
- Heat energy
- Light energy
- Sound energy .
- Electrical energy
- Nuclear energy

In this class we will discuss only kinetic energy and potential energy in detail. The ‘kinetic energy’ and ‘potential energy’ taken together is known as ‘mechanical energy’.

### Kinetic Energy

A moving cricket ball can do work in pushing back the stumps (see Figure); moving water can do work in turning a turbine for generating electricity; and moving wind can do work in turning the blades of wind-mill. Thus, a moving body is capable of doing work and hence possesses energy. The energy of a body due to its motion is called kinetic energy.

A moving hammer drives a nail into wood because of its kinetic energy and a moving bullet can penetrate even a steel plate due to its kinetic energy which it has on account of its high speed. In fact, every object around us which is moving possesses kinetic energy. In other words, every object around us which has speed, possesses kinetic energy.

For example, a runner has kinetic energy; a running motorcycle has kinetic energy ; a running car (or bus) has kinetic energy; a falling stone has kinetic energy ; and an arrow flying through the air has also kinetic energy. When a moving body is brought to rest (stopped) by an opposing force, the kinetic energy is lost, being used up to do work in overcoming the resistance of opposing force. We will now derive a formula for calculating the kinetic energy of a moving body.

### Formula for Kinetic Energy

The kinetic energy of a moving body is measured by the amount of work it can do before coming to rest. Suppose a body, such as a ball, of mass m and moving with a velocity v is at position A (see Figure). Let it enter into a medium M, such as air, which opposes the motion of the body with a constant force F.

As a result of the opposing force, the body will be constantly retarded, that is, its velocity decreases gradually and it will come to rest (or stop) at position B after travelling a distance s. So, the final velocity V of the body becomes zero.

(i) In going through the distance s against the opposing force F, the body has done some work. This work is given by:

Work = Force × Distance

or W = F × s

At position B the body is at rest, that is, it has no motion and hence no kinetic energy. This means that all the kinetic energy of the body has been used up in doing the work W. So, the kinetic energy must be equal to this work W. Thus,

Kinetic energy = W

or Kinetic energy = F × s …………….. (1)

(ii) If a body has an initial velocity V, final velocity ‘V’, acceleration ‘a’ and travels a distance ‘s’, then according to the third equation of motion :

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

(Please note that we have written V^{2} = v^{2} + 2as instead of the usual v^{2} – u^{2} + 2as. But it should not make any difference)

In the above example, we have:

Initial velocity of the body = v (Supposed)

Finalvelocity of the body, V= 0 (The body stops)

Acceleration = – a (Retardation)

and Distance travelled = s

Now, putting these values in the above equation, we get:

(0)^{2} = v^{2} – 2as

or v^{2} = 2as

From Newton’s second law of motion, we have :

F = m × a

or a = \(\frac{F}{m}\)

Putting this value of acceleration ‘a’ in equation (2), we get:

v^{2} = \(\frac{2 \times F \times s}{m}\)

or F × s = \(\frac{1}{2}\) mv^{2}

But from equation (1), F × s = Kinetic energy. So, comparing equations (1) and (3), we get:

Kinetic energy = \(\frac{1}{2}\) mv^{2}

where m = mass of the body

and v = velocity of the body (or speed of the body)

Thus, a body of mass m and moving with a velocity v has the capacity of doing work equal to \(\frac{1}{2}\) mv^{2} before it stops.

### Some Important Conclusions

We have just seen that the kinetic energy of a body of mass m and moving with a velocity (or speed) v is given by the formula :

Kinetic energy = \(\frac{1}{2}\) mv^{2}

From this formula, it is clear that:

- the kinetic energy of a body is directly proportional to the mass of the body, and
- the kinetic energy of a body is directly proportional to the square of velocity of the body (or square of the speed of the body).

Since the kinetic energy of a body is directly proportional to its mass, therefore, if the mass of a body is doubled, its kinetic energy also gets doubled and if the mass of a body is halved, its kinetic energy also gets halved (provided its velocity remains the same).

Again, since the kinetic energy of a body is directly proportional to the square of its velocity, therefore, if the velocity of a body is doubled, its kinetic energy becomes four times, and if the velocity of a body is halved, then its kinetic energy becomes one-fourth. It is obvious that doubling the velocity has a greater effect on the kinetic energy of a body than doubling its mass.

Since the kinetic energy of a body depends on its mass and velocity, therefore, heavy bodies moving with high velocities have more kinetic energy (they can do more work), than slow moving bodies of small mass. This is the reason why a blacksmith uses a heavier hammer than the one used by a goldsmith. It has been found that a driver increases the speed (velocity) of his car on approaching a hilly road.

Let us see why this is done. When a car is moving on a flat road, it has to do work to overcome the friction of the road and air resistance but no work is done against the force of gravity. On the other hand, when the car is going up the hill, then in addition to friction and air resistance, it has to do work against the force of gravity.

Thus, a driver increases the speed of his car on approaching a hilly road to give more kinetic energy to the car so that it may go up against gravity. Please note that in the above discussion on kinetic energy we have mostly used the term “velocity”. The term “speed” can also be used in place of “velocity” everywhere in the above description of kinetic energy.

Another point to be noted is that ‘Kinetic Energy’ is also denoted by the symbol K.E. A yet another symbol for kinetic energy is Ek (where E stands for Energy and k for kinetic). We will now solve some numerical problems based on kinetic energy.

**Example Problem 1.**

Calculate the kinetic energy of a body of mass 2 kg moving with a velocity of 0.1 metre per second.

**Solution:**

The formula for calculating kinetic energy is :

Kinetic energy = \(\frac{1}{2}\) mv^{2}

Here, Mass, m = 2 kg

And, Velocity, v = 0.1 m/s

So, putting these values in the above formula, we get:

Kinetic energy = \(\frac{1}{2}\) × 2 × (0.1)^{2}

= \(\frac{1}{2}\) × 2 × 0.1 × 0.1

= 0.01 J

**Example Problem 2.**

Two bodies of equal masses move with uniform velocities v and 3v respectively. Find the ratio of their kinetic energies.

**Solution:**

In this problem, the masses of the two bodies are equal, so let the mass of each body be m. We will now write down the expressions for the kinetic energies of both the bodies separately.

(i) Mass of first body = m

Velocity of first body = v

So, K.E. of first body = \(\frac{1}{2}\) mv^{2} ……… (1)

(ii) Mass of second body = m

Velocity of second body = 3v

So, K.E. of second body = \(\frac{1}{2}\) m (3v)^{2}

= \(\frac{1}{2}\) m × v^{2}

= \(\frac{9}{2}\) mv^{2} …………… (2)

Now, to find out the ratio of kinetic energies of the two bodies, we should divide equation (1) by equation (2), so that:

\(\frac{\text { K.E. of first body }}{\text { K.E. of second body }}\) = \(\frac{\frac{1}{2} m v^2}{\frac{9}{2} m v^2}\)

or \(\frac{\text { K.E. of first body }}{\text { K.E. of second body }}\) = \(\frac{1}{9}\)

Thus, the ratio of the kinetic energies is 1 : 9.

We can also write down the equation (3) as follows :

K.E. of second body = 9 × K.E. of first body

That is, the kinetic energy of second body is 9 times the kinetic energy of the first body. It is clear from this example that when the velocity (or speed) of a body is “tripled” (from v to 3v), then its kinetic energy becomes “nine times”.

**Example Problem 3.**

How much work should be done on a bicycle of mass 20 kg to increase its speed from 2 m s^{-1} to 5 m s^{-1}? (Ignore air resistance and friction).

**Solution:**

We know that whenever work is done, an equal amount of energy is used up. So, the work done in this case will be equal to the change in kinetic energy of bicycle when its speed changes from 2 m s^{-1} to 5 m s^{-1}.

(a) In the first case :

Mass of bicycle, m = 20 kg

Speed of bicycle, v = 2 m s^{-1}

So, Kinetic energy, E_{k} = \(\frac{1}{2}\) mv^{2}

= \(\frac{1}{2}\) × 20 × (2)^{2}

= 10 × 4

= 40 J

(b) In the second case :

Mass of bicycle, m = 20 kg

And, Speed of bicycle, v = 5 m s^{-1}

So, Kinetic energy, E_{k} = \(\frac{1}{2}\) mv^{-1}

= \(\frac{1}{2}\) × 20 × (5)^{2}

= 10 × 25

= 250 J

Now, Work done = Change in kinetic energy

= 250 – 40

= 210 J

Thus, the work done is 210 joules.

### Potential Energy

Suppose a brick is lying on the ground. It has no energy so it cannot do any work. Let us lift this brick to the roof of a house (see Figure). Now, some work has been done in lifting this brick against the force of gravity. This work gets stored up in the brick in the form of potential energy.

Thus, the energy of a brick lying on the roof of a house is due to its higher position with respect to the ground. And if this brick falls from the roof-top, it can do some (undesirable) work in breaking the window-panes or somebody’s head ! The energy of brick lying on the roof-top is known as gravitational potential energy because it has been acquired by doing work against gravity.

Another type of potential energy is elastic potential energy, which is due to a change in the shape of the body. The change in shape of a body can be brought about by compressing, stretching, bending or twisting. Some work has to be done to change the shape of a body (temporarily). This work gets stored in the deformed body in the form of elastic potential energy.

When this deformed body is released, it comes back to its original shape and size and the potential energy is given out in some other form. For example, a wound¬up circular spring possesses elastic potential energy which drives a wound-up toy (such as a toy car). Figure (a) shows the normal shape of the circular spring used in winding toys.

When we wind-up the spring of a toy-car by using a winding key, then some work is done by us due to which the spring gets coiled more tightly [see figure (b)]. The work done in winding the spring gets stored up in the tightly coiled-up spring (or wound-up spring) in the form of elastic potential energy. When the wound-up spring is slowly released, its potential energy is gradually converted into kinetic energy which turns the wheels of the toy car and makes it run.

Thus, a wound-up spring can do work in returning to its original shape during unwinding. The potential energy of a wound-up spring is not due to its position above the ground, it is due to the change in its shape. Let us take another example.

When we do work in stretching the rubber strings of a catapult (gulel), then the work done by us gets stored in the stretched rubber strings in the form of elastic potential energy [see Figure (a)]. The stretched strings of a catapult possess potential energy due to a change in their shape (because they become long and thin).

This energy of the stretched strings of the catapult can be used to throw away a piece of stone with a high speed [see Figure (b)]. We can now say that:

The energy of a body due to its position or change in shape is known as potential energy. Actually, the energy of a body due to its position above the ground is called gravitational potential energy and the energy of a body due to a change in its shape and size is called elastic potential energy. Elastic potential energy is associated with the state of ‘compression’ or ‘extension’ of an object.

For example, the energy possessed by a ‘compressed spring’ or an ‘extended spring’ (stretched spring) is the elastic potential energy. The gravitational potential energy as well as elastic potential energy are commonly known as just potential energy.

The water in a tank on the roof of a building possesses potential energy due to its position (height) above the ground. A stretched rubber band and compressed gas in a cylinder also possess potential energy but this is due to their change in shape or configuration.

A ceiling fan which has been switched off, water in the reservoir of a dam, a spring expanded beyond its normal shape, a rubber band lying on the table, and a stretched rubber band lying on the ground, all possess potential energy. A bent bow has also potential energy stored in it.

The potential energy stored in the bent bow (due to change in its shape) is used in the form of kinetic energy in throwing off an arrow. It is obvious that a body may possess energy even when it is not in motion. And this energy is called potential energy.

A body can have both potential energy as well as kinetic energy at the same time. The sum of the potential and kinetic energies of a body is called its mechanical energy. A flying bird, a flying aeroplane, and a man climbing a hill, all have kinetic energy as well as potential energy.

A stationary stone lying at the top of a hill has only potential energy. When this stone starts rolling downwards, it has both kinetic and potential energy. And when the stone reaches the bottom of the hill, it has only kinetic energy. We will now derive a formula for calculating the gravitational potential energy of a body.

### Formula for Potential Energy

The potential energy of a body is due to its higher position above the earth and it is equal to the work done on the body, against gravity, in moving the body to that position. So, to find out the potential energy of a body lying at a certain height, all that we have to do is to find out the work done in taking the body to that height.

Suppose a body of mass m is raised to a height h above the surface of the earth (see Figure). The force acting on the body is the gravitational pull of the earth m × g which acts in the downward direction. To lift the body above the surface of the earth, we have to do work against this force of gravity. Now,

Work done = Force × Distance

So, W = m × g × h

This work gets stored up in the body as potential energy. Thus,

Potential energy = m × g × h

where m = mass of the body

g = acceleration due to gravity

and h = height of the body above a reference point, say the surface of earth

We will now solve some problems based on potential energy. Please note that ‘Potential Energy’ is usually denoted by the letters P.E. Another symbol for potential energy is E_{p} (where E stands for Energy and p for potential).

**Example Problem 1.**

If acceleration due to gravity is 10 m/s^{2}, what will be the potential energy of a body of mass 1 kg kept at a height of 5 m ?

**Solution:**

The potential energy of a body is calculated by using the formula :

Potential energy = m × g × h

In this case, Mass, m = 1 kg

Acceleration due to gravity, g = 10 m/s^{2}

And, Height, h = 5 m

So, putting these values in the above formula, we get:

Potential energy = 1 × 10 × 5

= 50 J

Thus, the potential energy of the body is 50 joules.

**Example Problem 2.**

A bag of wheat weighs 200 kg. To what height should it be raised so that its potential energy may be 9800 joules ? (g = 9.8 m s^{-2})

**Solution:**

Here, Potential energy, P.E. = 9800 J

Mass, m = 200 kg

Acceleration due to gravity, g = 9.8 ms^{-2}

And, Height, h = ? (To be calculated)

Now, putting these values in the formula :

P.E. = m × g × h

We get: 9800 = 200 × 9.8 × h

So, h = \(\frac{9800}{200 \times 9.8}\)

h = 5 m

Thus, the bag of wheat should be raised to a height of 5 metres.