Get A+ with YaClass!

Register now to understand any school subject with ease and get great results on your exams!

### Theory:

In the previous sections, we discussed that one form of energy could be converted to another form.

**What results in the total energy of a system during or after the conversion process?**

Whenever energy gets converted, the total energy remains unchanged. This is the law of conservation of energy.

According to this law, energy can only be transformed from one form to another; it can neither be created nor destroyed. The total energy before and after the conversion remains the same.

The law of conservation of energy is valid in all situations and for all kinds of transformations.

Let us consider a simple example,

An object with mass, \(m\) is made to fall freely from a height, \(h\).

In the beginning,

**Potential energy of the object**\(=\) \(mgh\) and

**Kinetic energy of the object**\(=\) 0

**Why is the kinetic energy zero?**

It is zero because the velocity of the object is zero. The total energy of the object is thus \(mgh\).

As it falls,

The potential energy of the object will change into kinetic energy.

At a particular time, consider the velocity of the object is \(v\) ,

**Kinetic energy of the object**\(=\) $\frac{1}{2}\times \phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}{v}^{2}$ .

As the object continues to fall, the potential energy would decrease while the kinetic energy would increase.

When the object is about to reach the ground, \(h\) \(=\) \(0\) and \(v\) will be the highest. Therefore, kinetic energy would be the largest and potential energy the least. However, the sum of the object's potential energy and kinetic energy would be the same at all points.

That is,

Potential Energy \(+\) Kinetic Energy \(=\) Constant

or

$(m\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}g\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}h)\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}(\frac{1}{2}\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}{v}^{2})\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{Constant}$

The sum of kinetic energy and potential energy of an object is its total mechanical energy. We find that during the free fall of the object, the decrease in potential energy appears as an equal amount of increase in kinetic energy at any point in its path. (Here, the effect of air resistance on the motion of the object has been ignored.) There is thus a continual transformation of gravitational potential energy into kinetic energy.