Potential energy is energy that is already possessed by an object. This could be do to the objects position, current stresses that are within the object, or possibly an electrical charge.

One common form of potential energy is gravitational potential energy. Gravitational potential energy is dependent on the objects mass as well as distance between that object and another object such as the ground. To determine the gravitational potential energy of an object the equation below would be used.

**(Eq 1) ** $U_{1-2}=WΔy$

W = Weight

Δy = suspended height

U_{1-2} = Potential Energy

As seen in the image above as an object is hanging in space but has the potential to fall. It will have a potential energy that is the product of that objects weight and the distance the object is from the ground.

**Example**

A weight with a mass of 50kg is being held 2.5m above the ground. What amount of energy is stored in the object where the object has the potential to fall the full height and hit the ground?

**Solution**

$U_{1-2}=(9.81)(50)(2.5)=1.2~kJ$

Another example of potential energy is when an internal stress is placed on an object. An example of this would be compressing a spring. As the spring is compressed it will provide a force on the object that is compressing the spring. The further the spring is compressed the greater the force. This force is due to the energy being stored into the spring that will release once the object that is compressing the spring is removed. To determine the potential energy of a spring the following equation would be used.

**(Eq 2)** $U_{1-2}=-\left( \frac{1}{2}ks_2^2-\frac{1}{2}ks_1^2 \right)$

k = spring constant

s_{2} = Initial spring length

s_{1} = New spring length

The variable k in equation 2 represents that spring constant, which represent the amount of deflection caused by a certain amount of force. Refer to equation 3.

**(Eq 3)** $k=\frac{f}{δ}$

f = force

δ = change in length

The above equation can be applied to any object that is under a mechanical stress once the spring stiffness of that object is determined.

**Example**

A spring has a spring stiffness of 25$\frac{N}{mm}$. A force is deflecting the spring 20mm. How much energy is being stored in the spring?

**Solution**

$U_{1-2}=\frac{1}{2}(25)(20)^2=5~kJ$

## Kinetic Energy

Kinetic energy is the energy of an object due to the motion of that object. As an object accelerates a certain amount of work is required for that object to reach its new velocity. The work will cause a change in kinetic energy stored within the object. While on the other if an object was to remain at a constant velocity than its stored kinetic energy would remain constant. Finally, in order for an object to come to a complete rest all of its kinetic energy would need to be expelled.

An particles kinetic energy is based off its velocity as well as its mass as seen in the equation below.

**(Eq 4)** $T=\frac{1}{2}mv^2$

m = Mass

v = velocity

T = Kinetic Energy

**Example **

A particle is traveling at a constant velocity of 15$\frac{m}{s}$. The particle than accelerates to a new velocity of 22$\frac{m}{s}$. The particles mass is 25kg. What was the change in the particles kinetic energy as it velocity changed from 15$\frac{m}{s}$ to 22$\frac{m}{s}$? What is the particles total kinetic energy as 22$\frac{m}{s}$?

**Solution**

Determine the particle’s kinetic energy at 22$\frac{m}{s}$.

$T_2=\frac{1}{2}(25)(22)^2= 6.1~kJ$

Determine the particles kinetic energy at 15$\frac{m}{s}$.

$T_1=\frac{1}{2}(25)(15)^2= 2.8~kJ$

Determine the change in kinetic energy due to the change in velocity.

$T_{1-2}=6.1-2.8=3.3~kJ$