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What is Potential Energy?

  • Last Updated : 14 Jul, 2021
Geek Week

As a common notion, the word “energy” means the capacity to come into action. Whenever someone says that a particular person is energetic. This means that a particular person is eager and very capable of doing things. The term “potential” is also used to denote the possibility of something happening. These things relate to energy. Everything that is at rest under a force has the potential to be in action. This means potential energy is stored inside that object. 

Potential Energy

Let us consider a mass M that is kept on the ground. This mass M is taken to a height of “h”. In the process, the gravitational force acts on the mass and does work on it. The work done while going against gravity is stored in the block in terms of potential energy. When the block is released to fall freely, this potential energy decreases and gets converted into kinetic energy. 

Potential energy is the energy acquired by the body by the virtue of its position or configuration. 

In the above experiment, it was mentioned that the work done against the force of gravity is converted into potential energy. The gravitational force on the block will be, 

F = Mg



While rising through the height “h”. The work done on the block will be, 

W = F.d

⇒W = (Mg)h

⇒W = Mgh

Thus, this work done is converted into potential energy. 

P.E = W = Mgh

Since the force is gravity. This is called gravitational potential energy. In general, potential energy can be due to work done under a lot of different forces. For example, Electrostatic Force, Spring Restoring Force, etc. Since potential energy is also a type of energy. Its unit is Joules. 

Potential Energy of a Spring



Hooke’s law states how the restoring force in the spring varies as the net displacement from the mean position of the spring. Considering the net displacement to be \Delta x and the restoring force being denoted by F, 

F = -kx

For a variable force F, and the net displacement x, 

W = \int^{x}_{0}Fdx

Now at the displacement x, for an infinitesimally small-displacement \Delta x and force F, 

dW = Fdx

⇒ dW = -kxdx

Integrating the above equation for the total work done, 

dW = kxdx

⇒∫dW = ∫kxdx



⇒ W = \frac{kx^2}{2}

This work done is also stored as potential energy. Thus, potential energy stored in the spring due to displacement “x” will be,

P.E = \frac{kx^2}{2}

The notion of potential energy is only limited to the class of forces where work done gets converted into potential energy. For example, friction does not come in that category. In the case of friction, the work done is dissipated in the form of heat. 

General form of Potential Energy

In general, potential energy V(x) in the presence of force F(x) is defined as, 

F(x) = -\frac{dV(x)}{dx}

This implies that, 

\int^{x_f}_{x_i}F(x)dx = -\int^{v_f}_{v_i}{dV(x)}{dx} = V_i - V_f

Work done by such forces is dependent upon the initial and final values of x.  Such forces are called conservative forces. 



Sample Problems

Question 1: Find the potential energy of a ball of mass 3Kg kept at a height of 20m. 

Solution: 

Given: 

Mass “m” = 3Kg, height “h” = 20m. 

P.E = mgh 

Substituting the values in the given equation. 

⇒ P.E = (3)(10)(20) 

⇒ P.E = 600J

Question 2: Find the potential energy of a ball of mass 5Kg kept at a height of 10m. 

Solution: 

Given: 

Mass “m” = 5Kg, height “h” = 10m. 

P.E = mgh 

Substituting the values in the given equation. 

⇒ P.E = (5)(10)(10) 

⇒ P.E = 500J

Question 3: Find the potential energy stored in spring with a spring constant of 50N/m, when it is stretched for 0.2m. 

Solution: 

Given: 

K = 50 N/m and the \Delta x = 0.2m

Energy stored in a spring is given by, 

P.E = \frac{kx^2}{2}

Plugging these values in the equation, 

P.E = \frac{kx^2}{2}

⇒ P.E = \frac{(50)(0.2)^2}{2}

⇒ P.E = \frac{(50)(0.04)}{2}

⇒ P.E = \frac{2}{2}

⇒ P.E = 1J

Question 4: Find the potential energy stored in spring with a spring constant of 100N/m, when it is stretched for 0.5m. 

Solution: 



Given: 

K = 100 N/m and the \Delta x = 0.5m

Energy stored in a spring is given by, 

P.E = \frac{kx^2}{2}

Plugging these values in the equation, 

P.E = \frac{kx^2}{2}

⇒ P.E = \frac{(100)(0.5)^2}{2}

⇒ P.E = \frac{(100)(0.25)}{2}

⇒ P.E = \frac{25}{2}

⇒ P.E = 12.5J



Question 5: The potential energy of the block changes 50J when it is thrown upwards from a height of 10m. The mass of the block is 1Kg. Find the height which the block reaches. 

Solution: 

Let’s say the initial height and final height are, 

hi = 10m and hf = ? 

Given: m = 1Kg and \Delta P.E = 50J

ΔP.E=P.Ef​−P.Ei​=50

= mghf​−mghi​=50

=mg(hf​−hi​)=50

=(1)(10)(hf​−hi​)=5

=hf​−hi​=5

=hf​−10

=15 = hf

The final height will be 15m. 

Question 6: A block connected to a spring was compressed to 0.2m. The mass of the block is 2Kg. Find the velocity of the block when it reaches its natural position. Given. K = 100N/m.

Solution: 

Given: 

K = 100 N/m and the \Delta x = 0.2m

Energy stored in a spring is given by, 

P.E = \frac{kx^2}{2}

Plugging these values in the equation, 

P.E = \frac{kx^2}{2}

⇒ P.E = \frac{(100)(0.2)^2}{2}

⇒ P.E = \frac{(100)(0.04)}{2}

⇒ P.E = 2J

Now, when the spring is released, it transfers its potential energy to the block, and it converts into the kinetic energy of the block. 

K.E = \frac{1}{2}mv^2

⇒ 2 = \frac{1}{2}2v^2

⇒ √2 = v

v = 1.414 m/s. 

Question 7: A block connected to a spring was compressed to 0.5m. The mass of the block is 1Kg. Find the velocity of the block when it reaches its natural position. Given. K = 100N/m



Solution: 

Given: 

k = 100 N/m and the \Delta x = 0.5m

Energy stored in a spring is given by, 

P.E = \frac{kx^2}{2}

Plugging these values in the equation, 

P.E = \frac{kx^2}{2}

⇒ P.E = \frac{(100)(0.5)^2}{2}

⇒ P.E = \frac{25}{2}

⇒ P.E = 12.5J

Now, when the spring is released, it transfers its potential energy to the block, and it converts into the kinetic energy of the block. 

K.E = \frac{1}{2}mv^2

⇒ 12.5 = \frac{1}{2}1v^2

⇒ √25 = v

v = 5 m/s. 

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