In the previous section we saw basic details about potential energy. In this section we will see the Law of conservation of Energy.
In our day to day life, we see different sources of energy. Let us see some examples:
Example 1: An ordinary torch cell is a source of energy. In it, energy is stored in the form of chemical energy. But we cannot use this chemical energy directly. In the torch, the chemical energy is first converted into electrical energy. This electrical energy is then converted into light energy.
Example 2: In a motor car, we fill petrol. This petrol contains lot of chemical energy. In the car’s engine, this petrol burns, and causes the engine parts to move. The movement of engine parts is transmitted to the wheels of the car. Thus the car moves forward. So the chemical energy in petrol gets converted into kinetic energy of the car.
Example 3: The sun provides heat energy. This energy causes the water in oceans and lakes to evoporate. The vapours thus formed will merge together to form clouds. When the clouds cool, they come down as rain. This rain is collected in reservoirs. The reservoirs are constructed at higher levels. So the water in them have high potential energy. This water flows down through special pipes with great force and turn the turbines of generators, thus producing electricity. So we find that heat energy of the sun is first converted into potential energy of water in the reservoirs. This potential energy is then converted into electrical energy. We can say that, the heat energy from the sun got finally converted into electrical energy
Example 4: The food that we eat contains a lot of chemical energy. We get energy for our daily activities from this chemical energy. When we climb the stairs of a building and reach an upper floor, our muscles have to do a lot of work. Much energy that we received from the food will be used up when we climb steps. But that energy will not be wasted because, when we reach a higher level, our potential energy increases. We can say that the chemical energy from the food, got finally converted into potential energy.
Example 1. In the case of torch cells, the chemical energy got converted into light energy.
• According to the law, total energy derived from the cell is equal to the total light energy produced.
• But such a perfect conversion is not possible. Because, a portion of the chemical energy is lost as heat energy.
• But if we calculate the sum, we will find that the law is valid. That is.,
• Chemical energy derived from the cell =
[Light energy] + [Heat energy wasted]
Example 2: Chemical energy derived from the petrol
= [Kinetic energy of the car] + [Heat energy wasted in the engines] + [Energy lost to overcome friction between tyres and road when the car moves] + [Energy lost to overcome air resistance encountered when the car moves]
Example 3: Heat energy derived from the sun =
[Electrical energy produced in the generators] + [Heat energy wasted in the generators and turbines] + [Energy lost to overcome friction between water and inside surface of pipes] + [Potential energy lost due to leakage in pipes]
Example 4: Chemical energy derived from the food =
[Potential energy of the person] + [Heat energy wasted in the body]
• So we see that the converted energy consists of various components. We will learn about those components in detail, in higher classes. At present, all we need to know is that, when we add those components, we will find that the total energy after conversion remains the same.
• In other words, the total energy before conversion is equal to the total energy after conversion.
The law of conservation of energy states that:
■ Energy can only be converted from one form to another; it can neither be created or destroyed.
In this chapter, we learned about kinetic energy and potential energy in some detail. So we will learn the details about the conversion between them. We will learn it with the help of an example.
1. An object A of mass of 25 kg is at rest. It is situated at a height of 6 m above the ground. Let this state of rest be 'Stage 0'. It is shown in fig.4.5 below:
• Let the acceleration due to gravity be taken as 10 m s-2.
• At this stage, it’s potential energy Ep is mgh = 25 × 10 × 6 = 1500 J
• At this stage, it’s kinetic energy Ek = zero. Because it has no velocity. It is at rest.
• So we can say that, at stage 0, total energy, Ep + Ek = 1500 + 0 = 1500 J
2. It is then allowed to fall freely. Consider the 'stage 1' when it travels 1 m from start .
• That is., At stage 1, the object is at a height of 5 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 5 = 1250 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 1 = 1500 - 1250 = 250 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×1 ⇒ v2 = 20 [u = 0 because, the object falls from rest]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 20 = 250 J
♦ So gain in kinetic energy = Ek at stage 1 - Ek at stage 0 = 250 - 0 = 250 J
• Also we can say that, at stage 1, total energy, Ep + Ek = 1250 + 250 = 1500 J
3. Consider the stage 2 when it travels 2 m from start .
• That is., At stage 2, the object is at a height of 4 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 4 = 1000 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 2 = 1500 - 1000 = 500 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×2 ⇒ v2 = 40 [Note that in this problem, we do not need to calculate the square root to find the actual 'v'. Because, we will be using 'v2' in the next step]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 40 = 500 J
♦ So gain in kinetic energy = Ek at stage 2 - Ek at stage 0 = 500 - 0 = 500 J
• Also we can say that, at stage 2, total energy, Ep + Ek = 1000 + 500 = 1500 J
4. Consider the stage 3 when it travels 3 m from start .
• That is., at stage 3, the object is at a height of 3 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 3 = 750 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 3 = 1500 - 750 = 750 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×3 ⇒ v2 = 60
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 60 = 750 J
♦ So gain in kinetic energy = Ek at stage 3 - Ek at stage 0 = 750 - 0 = 750 J
• Also we can say that, at stage 3, total energy, Ep + Ek = 750 + 750 = 1500 J
5. Consider the stage 4 when it travels 4 m from start .
• That is., at stage 4, the object is at a height of 2 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 2 = 500 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 4 = 1500 - 500 = 1000 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×4 ⇒ v2 = 80
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 80 = 1000 J
♦ So gain in kinetic energy = Ek at stage 4 - Ek at stage 0 = 1000 - 0 = 1000 J
• Also we can say that, at stage 4, total energy, Ep + Ek = 500 + 1000 = 1500 J
6. Consider the stage 5 when it travels 5 m from start .
• That is., at stage 5, the object is at a height of 1 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 1 = 250 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 5 = 1500 - 250 = 1250 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×5 ⇒ v2 = 100
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 100 = 1250 J
♦ So gain in kinetic energy = Ek at stage 5 - Ek at stage 0 = 1250 - 0 = 1250 J
• Also we can say that, at stage 5, total energy, Ep + Ek = 250 + 1250 = 1500 J
7. Consider the stage 6 when it travels 6 m from start .
• That is., at stage 6, the object is at a height of 0 m from the ground.
♦ That means the object has reached the ground. When it reaches the ground, it's velocity will be zero. So we will consider the instant when it just reaches the ground. At this stage, the velocity will be the maximum.
• At this stage, it’s potential energy = mgh = 25 × 10 × 0 = 0 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 6 = 1500 - 0 = 1500 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×6 ⇒ v2 = 120
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 120 = 1500 J
♦ So gain in kinetic energy = Ek at stage 6 - Ek at stage 0 = 1500 - 0 = 1500 J
• Also we can say that, at stage 6, total energy, Ep + Ek = 0 + 1500 = 1500 J
In the next section, we will see Power.
In our day to day life, we see different sources of energy. Let us see some examples:
Example 1: An ordinary torch cell is a source of energy. In it, energy is stored in the form of chemical energy. But we cannot use this chemical energy directly. In the torch, the chemical energy is first converted into electrical energy. This electrical energy is then converted into light energy.
Example 2: In a motor car, we fill petrol. This petrol contains lot of chemical energy. In the car’s engine, this petrol burns, and causes the engine parts to move. The movement of engine parts is transmitted to the wheels of the car. Thus the car moves forward. So the chemical energy in petrol gets converted into kinetic energy of the car.
Example 3: The sun provides heat energy. This energy causes the water in oceans and lakes to evoporate. The vapours thus formed will merge together to form clouds. When the clouds cool, they come down as rain. This rain is collected in reservoirs. The reservoirs are constructed at higher levels. So the water in them have high potential energy. This water flows down through special pipes with great force and turn the turbines of generators, thus producing electricity. So we find that heat energy of the sun is first converted into potential energy of water in the reservoirs. This potential energy is then converted into electrical energy. We can say that, the heat energy from the sun got finally converted into electrical energy
Example 4: The food that we eat contains a lot of chemical energy. We get energy for our daily activities from this chemical energy. When we climb the stairs of a building and reach an upper floor, our muscles have to do a lot of work. Much energy that we received from the food will be used up when we climb steps. But that energy will not be wasted because, when we reach a higher level, our potential energy increases. We can say that the chemical energy from the food, got finally converted into potential energy.
• Thus we see a lot of energy conversions taking place around us.
• But whenever such conversions take place, a peculiar phenomenon occurs.
• That is., the total energy remains unchanged. This is called the Law of conservation of energy.
• Let us see how this law applies to the above examples:
Example 1. In the case of torch cells, the chemical energy got converted into light energy.
• According to the law, total energy derived from the cell is equal to the total light energy produced.
• But such a perfect conversion is not possible. Because, a portion of the chemical energy is lost as heat energy.
• But if we calculate the sum, we will find that the law is valid. That is.,
• Chemical energy derived from the cell =
[Light energy] + [Heat energy wasted]
Example 2: Chemical energy derived from the petrol
= [Kinetic energy of the car] + [Heat energy wasted in the engines] + [Energy lost to overcome friction between tyres and road when the car moves] + [Energy lost to overcome air resistance encountered when the car moves]
Example 3: Heat energy derived from the sun =
[Electrical energy produced in the generators] + [Heat energy wasted in the generators and turbines] + [Energy lost to overcome friction between water and inside surface of pipes] + [Potential energy lost due to leakage in pipes]
Example 4: Chemical energy derived from the food =
[Potential energy of the person] + [Heat energy wasted in the body]
• So we see that the converted energy consists of various components. We will learn about those components in detail, in higher classes. At present, all we need to know is that, when we add those components, we will find that the total energy after conversion remains the same.
• In other words, the total energy before conversion is equal to the total energy after conversion.
The law of conservation of energy states that:
■ Energy can only be converted from one form to another; it can neither be created or destroyed.
In this chapter, we learned about kinetic energy and potential energy in some detail. So we will learn the details about the conversion between them. We will learn it with the help of an example.
1. An object A of mass of 25 kg is at rest. It is situated at a height of 6 m above the ground. Let this state of rest be 'Stage 0'. It is shown in fig.4.5 below:
 |
| Fig.4.5 |
• At this stage, it’s potential energy Ep is mgh = 25 × 10 × 6 = 1500 J
• At this stage, it’s kinetic energy Ek = zero. Because it has no velocity. It is at rest.
• So we can say that, at stage 0, total energy, Ep + Ek = 1500 + 0 = 1500 J
2. It is then allowed to fall freely. Consider the 'stage 1' when it travels 1 m from start .
• That is., At stage 1, the object is at a height of 5 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 5 = 1250 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 1 = 1500 - 1250 = 250 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×1 ⇒ v2 = 20 [u = 0 because, the object falls from rest]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 20 = 250 J
♦ So gain in kinetic energy = Ek at stage 1 - Ek at stage 0 = 250 - 0 = 250 J
• Also we can say that, at stage 1, total energy, Ep + Ek = 1250 + 250 = 1500 J
3. Consider the stage 2 when it travels 2 m from start .
• That is., At stage 2, the object is at a height of 4 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 4 = 1000 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 2 = 1500 - 1000 = 500 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×2 ⇒ v2 = 40 [Note that in this problem, we do not need to calculate the square root to find the actual 'v'. Because, we will be using 'v2' in the next step]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 40 = 500 J
♦ So gain in kinetic energy = Ek at stage 2 - Ek at stage 0 = 500 - 0 = 500 J
• Also we can say that, at stage 2, total energy, Ep + Ek = 1000 + 500 = 1500 J
4. Consider the stage 3 when it travels 3 m from start .
• That is., at stage 3, the object is at a height of 3 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 3 = 750 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 3 = 1500 - 750 = 750 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×3 ⇒ v2 = 60
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 60 = 750 J
♦ So gain in kinetic energy = Ek at stage 3 - Ek at stage 0 = 750 - 0 = 750 J
• Also we can say that, at stage 3, total energy, Ep + Ek = 750 + 750 = 1500 J
5. Consider the stage 4 when it travels 4 m from start .
• That is., at stage 4, the object is at a height of 2 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 2 = 500 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 4 = 1500 - 500 = 1000 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×4 ⇒ v2 = 80
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 80 = 1000 J
♦ So gain in kinetic energy = Ek at stage 4 - Ek at stage 0 = 1000 - 0 = 1000 J
• Also we can say that, at stage 4, total energy, Ep + Ek = 500 + 1000 = 1500 J
6. Consider the stage 5 when it travels 5 m from start .
• That is., at stage 5, the object is at a height of 1 m from the ground.
• At this stage, it’s potential energy = mgh = 25 × 10 × 1 = 250 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 5 = 1500 - 250 = 1250 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×5 ⇒ v2 = 100
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 100 = 1250 J
♦ So gain in kinetic energy = Ek at stage 5 - Ek at stage 0 = 1250 - 0 = 1250 J
• Also we can say that, at stage 5, total energy, Ep + Ek = 250 + 1250 = 1500 J
7. Consider the stage 6 when it travels 6 m from start .
• That is., at stage 6, the object is at a height of 0 m from the ground.
♦ That means the object has reached the ground. When it reaches the ground, it's velocity will be zero. So we will consider the instant when it just reaches the ground. At this stage, the velocity will be the maximum.
• At this stage, it’s potential energy = mgh = 25 × 10 × 0 = 0 J
♦ So loss of potential energy = Ep at stage 0 - Ep at stage 6 = 1500 - 0 = 1500 J
• At this stage, we want to know it’s kinetic energy.
• For that, we want it’s velocity at this stage. As before, using the third equation of motion, we can calculate the velocity:
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×6 ⇒ v2 = 120
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 25 × v2 = 1⁄2 × 25 × 120 = 1500 J
♦ So gain in kinetic energy = Ek at stage 6 - Ek at stage 0 = 1500 - 0 = 1500 J
• Also we can say that, at stage 6, total energy, Ep + Ek = 0 + 1500 = 1500 J
We can write the above results in a tabular form as shown below:
From the above table, we can note the following points:
• At any stage, the loss of potential energy is equal to the gain in kinetic energy
• That is., what ever potential energy is lost, is gained by the kinetic energy
• So, there is a continuous transformation of energy from one form (potential) to another (kinetic)
• The total energy, that is., the sum Ep + Ek is always a constant
■ So this is an excellent example to prove the Law of conservation of energy
Now we will see a solved example
Solved example 4.8
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. Take g = 10 m s-2
Solution:
Mass m = 40 kg, Height h = 5 m
Part (i): Ep = mgh = 40 × 10 × 5 = 2000 J
Part (ii): We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage. We want the velocity when it is 2.5 m from the ground.
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×2.5 ⇒ v2 = 50 [u = 0 because, the object falls from rest]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 40 × v2 = 1⁄2 × 40 × 50 = 1000 J
Mass m = 40 kg, Height h = 5 m
Part (i): Ep = mgh = 40 × 10 × 5 = 2000 J
Part (ii): We know that, when an object is allowed to fall freely, it accelerates towards the earth, and so it’s velocity increases. Using the third equation of motion, we can calculate the velocity at any stage. We want the velocity when it is 2.5 m from the ground.
• v2 = u2 + 2as ⇒ v2 = 0 + 2gs ⇒ v2 = 0 + 2×10×2.5 ⇒ v2 = 50 [u = 0 because, the object falls from rest]
• So the kinetic energy, Ek = 1⁄2 × mv2 = 1⁄2 × 40 × v2 = 1⁄2 × 40 × 50 = 1000 J
In the next section, we will see Power.
