In the previous section we saw basic details about kinetic energy. In this section we will see potential energy.
■ Take a rubber band. Hold it's one end and pull the other end.
• The band stretches. In this stretched position, the band will try to pull our fingers at the two ends. We can feel it because, we have to apply force to keep the band in a stretched position. So it is clear that the band is applying a pulling force.
• From where did it get the energy to apply this force?
• We know that a rubber band which is not stretched, will not be able to apply any force.
• So it is clear that, the 'process of stretching' gave it some energy. The energy that we applied to stretch the rubber band, got stored in it. It will be stored in it until it regains it's original shape.
■ Similar is the case of bow and arrow. When the string of the arrow is tightened, the bow bends and is stretched. The energy used to stretch the bow, gets stored in it.
• When the arrow is placed in the bow and pulled back wards, the bow bends further and more energy gets stored.
• When the arrow is released, the bow regains it's original position. When this happens, the stored energy is released. This energy is converted into kinetic energy of the arrow and so the arrow moves forward.
■ Consider the stone raised to a height above ground. It possess some energy. It can now do some work, like driving a nail into a wooden piece.
• It acquired energy because some work was done to raise it to higher position.
• A stone at ground level will not be able to do work.
■ In the above three cases, the objects rubber band, bow and stone acquired energy because some work was done on them.
■ The work done got stored in them as potential energy.
We have to note some peculiarities of potential energy:
• The 'work done on an object' is stored as potential energy.
• But we saw that kinetic energy is also equal to the 'work done on the object'.
■ So what is the difference?
• In kinetic energy, the work done causes the object to move continuously with a certain velocity
• But in potential energy, there is no continuous movement for the object. There is only 'change in configuration' or 'change in position'
♦ In the case of rubber band, the band changed to a stretched configuration
♦ In the case of bow, it changed to a stretched configuration
♦ In the case of stone, it's position changed from ground level to a higher level
■ The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.
1. Consider an object of mass m. We have seen that it will have a weight W = mg. Where g is the acceleration due to gravity. (Details here)
The earth is pulling this object towards it's centre. The force of this pull is mg newton.
2. So we have to apply an equal and opposite force to raise the object above the ground. That is., we must apply a force of mg in the upward direction.
3. We know that work done = force × displacement. If the object is raised to a height h, the displacement is h. So work done on the object = mgh.
4. This much work gets stored in the object as potential energy. It is called gravitational potential energy.
5. So, the gravitational potential energy of an object of mass m situated at a height h is equal to mgh. We can write it in the form of an equation:
Eq.4.2:
Gravitational potential energy Ep = mgh
We have to note two important points while considering gravitational potential energy:
First point:
• Consider 'object A' with mass m, remaining at ground level in fig.4.3 below. It cannot do any work on another object which is at the same ground level.
• So, if we take the ground level as our 'zero level' or 'datum level':
♦ The object A at height 'h' above datum will have an Ep equal to mgh
♦ The object A at the ground level is 'useless' as far as potential energy is concerned
• But, if there is a basement level at a height h1 below the ground level, the object A at ground level is 'not useless'. It can do a work equal to mgh1 on an object at the basement level.
■ So it is important to specify a datum when we use potential energy
Second point:
• Consider fig.4.4 below. In the first case, the object is taken to a height h above the ground level, through a straight line path. This is indicated as 'path 1'.
• In the second case, the object is taken through 'path 2' which is a zigzag path.
• The final position in both cases is 'at a height h above the ground level'.
• So the final potential energy in both cases will be the same: mgh.
■ So we can say that, the potential energy depends on the height from the datum level. It does not depend on the path taken to reach the height.
Now we will see some solved examples
Solved example 4.6
Find the energy possessed by an object of mass 12 kg when it is at a height of 4 m above the ground. Given, g = 9.8 m s-2.
Solution:
• Mass of the object = 12 kg
• Height above the ground at which the object is situated = 4 m
• Given g = 9.8 m s-2.
• We have: Ep = mgh = 12 × 9.8 × 4 = 470.4 J
Solved example 4.7
An object of mass 15 kg is at a certain height above the ground. If the potential energy of the object is 900 J, find the height at which the object is with respect to the ground. Given, g = 10 m s-2.
Solution:
• Mass of the object = 15 kg
• Potential energy of the object = 900 J
• Given g = 10 m s-2.
• We have: Ep = mgh ⇒ 900 = 15 × 10 × h ⇒900 = 150h ⇒h = 6 m
In the next section, we will see Law of Conservation of Energy.
■ Take a rubber band. Hold it's one end and pull the other end.
• The band stretches. In this stretched position, the band will try to pull our fingers at the two ends. We can feel it because, we have to apply force to keep the band in a stretched position. So it is clear that the band is applying a pulling force.
• From where did it get the energy to apply this force?
• We know that a rubber band which is not stretched, will not be able to apply any force.
• So it is clear that, the 'process of stretching' gave it some energy. The energy that we applied to stretch the rubber band, got stored in it. It will be stored in it until it regains it's original shape.
■ Similar is the case of bow and arrow. When the string of the arrow is tightened, the bow bends and is stretched. The energy used to stretch the bow, gets stored in it.
• When the arrow is placed in the bow and pulled back wards, the bow bends further and more energy gets stored.
• When the arrow is released, the bow regains it's original position. When this happens, the stored energy is released. This energy is converted into kinetic energy of the arrow and so the arrow moves forward.
■ Consider the stone raised to a height above ground. It possess some energy. It can now do some work, like driving a nail into a wooden piece.
• It acquired energy because some work was done to raise it to higher position.
• A stone at ground level will not be able to do work.
■ In the above three cases, the objects rubber band, bow and stone acquired energy because some work was done on them.
■ The work done got stored in them as potential energy.
We have to note some peculiarities of potential energy:
• The 'work done on an object' is stored as potential energy.
• But we saw that kinetic energy is also equal to the 'work done on the object'.
■ So what is the difference?
• In kinetic energy, the work done causes the object to move continuously with a certain velocity
• But in potential energy, there is no continuous movement for the object. There is only 'change in configuration' or 'change in position'
♦ In the case of rubber band, the band changed to a stretched configuration
♦ In the case of bow, it changed to a stretched configuration
♦ In the case of stone, it's position changed from ground level to a higher level
■ The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.
Gravitational potential energy
We have seen that, when an object is raised to a height, potential energy gets stored in that object. Let us calculate how much joules of energy is stored in this way:1. Consider an object of mass m. We have seen that it will have a weight W = mg. Where g is the acceleration due to gravity. (Details here)
The earth is pulling this object towards it's centre. The force of this pull is mg newton.
2. So we have to apply an equal and opposite force to raise the object above the ground. That is., we must apply a force of mg in the upward direction.
3. We know that work done = force × displacement. If the object is raised to a height h, the displacement is h. So work done on the object = mgh.
4. This much work gets stored in the object as potential energy. It is called gravitational potential energy.
5. So, the gravitational potential energy of an object of mass m situated at a height h is equal to mgh. We can write it in the form of an equation:
Eq.4.2:
Gravitational potential energy Ep = mgh
We have to note two important points while considering gravitational potential energy:
First point:
• Consider 'object A' with mass m, remaining at ground level in fig.4.3 below. It cannot do any work on another object which is at the same ground level.
 |
| Fig.4.3 |
♦ The object A at height 'h' above datum will have an Ep equal to mgh
♦ The object A at the ground level is 'useless' as far as potential energy is concerned
• But, if there is a basement level at a height h1 below the ground level, the object A at ground level is 'not useless'. It can do a work equal to mgh1 on an object at the basement level.
■ So it is important to specify a datum when we use potential energy
Second point:
• Consider fig.4.4 below. In the first case, the object is taken to a height h above the ground level, through a straight line path. This is indicated as 'path 1'.
 |
| Fig.4.4 |
• The final position in both cases is 'at a height h above the ground level'.
• So the final potential energy in both cases will be the same: mgh.
■ So we can say that, the potential energy depends on the height from the datum level. It does not depend on the path taken to reach the height.
Now we will see some solved examples
Solved example 4.6
Find the energy possessed by an object of mass 12 kg when it is at a height of 4 m above the ground. Given, g = 9.8 m s-2.
Solution:
• Mass of the object = 12 kg
• Height above the ground at which the object is situated = 4 m
• Given g = 9.8 m s-2.
• We have: Ep = mgh = 12 × 9.8 × 4 = 470.4 J
Solved example 4.7
An object of mass 15 kg is at a certain height above the ground. If the potential energy of the object is 900 J, find the height at which the object is with respect to the ground. Given, g = 10 m s-2.
Solution:
• Mass of the object = 15 kg
• Potential energy of the object = 900 J
• Given g = 10 m s-2.
• We have: Ep = mgh ⇒ 900 = 15 × 10 × h ⇒900 = 150h ⇒h = 6 m
In the next section, we will see Law of Conservation of Energy.
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