Potential Energy

Potential energy is the energy stored in a body due to its position or configuration. It is produced when work is done against a conservative force. This stored energy has the ability to do work and can be converted into other forms of energy.

Examples:

• A book kept on a shelf has potential energy due to its height.
• Water stored in a dam possesses potential energy.
• A roller coaster at the highest point of the track has potential energy.

Formula

The formula for potential energy depends on the force acting on the object.

P.E = W = m × g × h

• m = mass of the object (in kilograms)
• g = acceleration due to gravity (9.8 m/s²).
• h = height of the object above the reference point (in meters)

S.I Unit: Potential energy has the same units as kinetic energy kg m²/s². All forms of energy are measured in Joules (J).

Types of Potential Energy

1. Gravitational Potential Energy

When a body is lifted vertically upward, work is done against the gravitational force acting on it. This work done is stored in the body as gravitational potential energy.

The gravitational force acting on the mass is F = Mg

The work done in lifting the mass through height h is:

W = F.h

⇒W = (Mg)h

⇒W = Mgh

This work is stored as potential energy: P.E = W = Mgh

Since this potential energy arises due to gravity,it 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. 

2. 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: 

Given:

Initial height: h_i = 10m

mass, m = 1 Kg

Potential energy, ΔP.E = 50 J

Acceleration due to gravity, g = 10 m/s²

ΔP.E = P.Ef ​− P.Ei​

ΔP.E = mghf​ − mghi​

Substitute the values

50 = (1) (10) hf ​− (1) (10) (10)

50 =10h_f − 100

Now Solve for h_f

10h_f = 50 + 100

10h_f​=50+100

h_f=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 1 kg. Find the velocity of the block when it reaches its natural position. Given. K = 100 N/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. 

Unsolved Problem

Question 1: A ball of mass 4 kg is thrown vertically upward from the ground with a speed such that it reaches a height of 15 m. Find the potential energy of the ball at the highest point. Take g = 10 m/s².

Question 2: A spring with spring constant k = 200 N/m is stretched by 0.3 m. Find the potential energy stored in the spring.

Question 3: A block of mass 2 kg is attached to a spring (spring constant k=150 N/m) and compressed by 0.4 m. The block is then released. Find the velocity of the block when it passes through the natural (unstretched) position.

Question 4: A ball of mass 5 kg is thrown upward from a height of 2 m with an initial speed such that it gains an additional 18 J of potential energy. Find the maximum height reached by the ball. Use g=10 m/s².

Question 5: A block of mass 1 kg is attached to a horizontal spring (k = 100 N/m) on a friction less surface. The spring is compressed by 0.5 m and then released. Find the kinetic energy of the block when the spring reaches its natural length.

Question 6: A block of mass 3 kg is lifted from the ground to a shelf of height 2.5 m. Find the work done against gravity and the potential energy of the block at that height.