Potential Energy


What is Potential Energy?

As we know, an object can store energy as the result of its position. In the case of a bow and an arrow, when the bow is drawn, it stores some amount of energy, which is responsible for the kinetic energy it gains, when it is released. Similarly, in the case of a spring, when it is displaced from its equilibrium position, it gains some amount of energy which we observe in the form of stress we feel in our hand upon stretching it. We can define potential energy as a form of energy that results from the alteration of its position or state.

Gravitational Potential Energy

Gravitational potential energy of an object is defined as the energy possessed by an object rose to a certain height against gravity. We shall formulate gravitational energy with the following example.

  • Consider an object of mass = m.
  • Placed at a height h from the ground, as shown in the figure.

Now, as we know, the force required to raise the object is equal to m×g of the object.

Potential Energy

As the object is raised against the force of gravity, some amount of work (W) is done on it.

Work done on the object = force × displacement.

So,

W = m×g×h = mgh

As per the law of conservation of energy, since the work done on the object is equal to m×g×h, the energy gained by the object = m×g×h, which in this case is the potential energy E.

E of an object raised to a height h above the ground = m×g×h

Potential Energy Example

It is important to note that, the gravitational energy does not depend upon the distance travelled by the object, but the displacement i.e., the difference between the initial and the final height of the object. Hence, the path along which the object has reached the height is not taken into consideration. In the example shown below, the gravitational potential energy for both the blocks A and B will be the same.

Example of Potential Energy

Question:

What will be the gravitational potential energy possessed by a ball of mass 1 kg when it is raised to a height of 6 m above the ground. (g = 9.8 m s–2)

Solution:

Here, mass of the object (m) = 1 kg,

Displacement (height) (h) = 10 m,

Acceleration due to gravity (g) = 9.8 m s–2.

Potential energy (p) = m×g×h = 1 kg × 9.8 m s–2 × 10 m = 98 J.


Practise This Question

An electric dipole of moment p is placed in the position of stable equilibrium in uniform electric field of intensity E. If it is rotated through an angle θ from the initial position, the potential energy of electric dipole in the final position is