Energy Calculations

Work Done

In order to move an object, it must be given Energy. This Energy is transferred to the object when a Force is applied to it. The process of transferring Energy in this way is referred to as 'Doing Work' on the object. 

The Work Done moving an object can be found using the following formula:-

Ew  =  F x d  


Ew  =  Work Done (J) 

F  =  Force (N) 

d  =  Distance the Force is applied over (m) 

Example 1 - 

A box is pushed horizontally a distance of 3 m. If a Force of 18 N was required, what is the Work Done in moving the box?

Ew  =  F x d

Ew  =  18 x 3

Ew  =  54 J 

Gravitational Potential Energy

The above formula can be applied when moving an object in any direction. If the object is moved vertically, a special variation of the formula can be used. 

For example, when lifting a box upwards, the person lifting must work against the Force of Gravity. The Work Done to the box is stored within it as Gravitational Potential Energy. 

If the box is dropped, then the stored energy is converted into Kinetic Energy as it falls. The Force required to lift an object is equal the force of Gravity on that object, which is more commonly known as its Weight ( see Applications of Forces - Unit 3 ). The Force required can be found using the formula below :-

W  =  m x g 

Where :-

W  =  Force due to Gravity (N) 

m  =  Mass (kg) 

g  =  Gravitational Field Strength (Nm-1

By substituting the Weight formula above into the Work Done formula, a formula for Gravitational Potential Energy can be found:-

Ep  =  m x g x h


Ep  =  Gravitational Potential Energy (J) 

m  =  Mass (kg) 

g  =   Gravitational Field Strength - 9.8 on Earth  (Nkg-1

h =  height (m) 

Example 2 - 

The olympic record for men's weightlifting is a total mass of 263 kg. If the record holder Hossein Rezazadeh lifts the mass to a height of 2.6 m, what is the Gravitational Energy gained by the mass?

Ep  =  mgh

Ep  =  263 x 9.8 x 2.6

Ep  =  6701 J 

Kinetic Energy 

When a Force is applied to an object, Work is Done upon it. This Energy is converted in the Energy of movement - Kinetic Energy. The Kinetic Energy of an object can be found using the following formula:-

Ek  =  1/2 x m x v2

Where :-

Ek  =  Kinetic Energy (J) 

m  =  Mass (kg) 

v  =  Velocity (ms-1

Note - In the above formula, it is only the velocity that is squared, not the whole formula.

Example 3 - 

In the film 'Back to the Future' Marty must drive the DeLorean at 88mph in order to travel through Time. What is the Kinetic Energy required for the DeLorean's Time Drive to work?

Mass of a DeLorean = 1233 kg

1mph = 0.447 ms-1

Ek  =  1/2mv2

Ek  =  0.5 x 1233 x ( 88 x 0.447 )2

Ek  =  954 x103

Ek  =  954 kJ 

Conservation of Energy Case Study : A Simple Pendulum 

The classic example of the conservation of Energy is a simple pendulum. A simple pendulum is a mass attached to a light string which is suspended as shown in the diagram below:-

When the Pendulum is pulled to one side, it is given Gravitational Potential Energy caused by the increase in height. 

When it is released, this Gravitational Potential Energy is converted into Kinetic Energy. 

At the bottom of the swing, all the Gravitational Potential energy has been converted to Kinetic Energy, and the Pendulum reaches a top speed. 

The Pendulum then slows down again as the Kinetic Energy is converted back to Gravitational Potential Energy, until it reaches a maximum height on the other side.

The animation below shows this process happening:-

As the Total Energy of the System must be conserved, it is possible to define the Total Energy as:-

ETotal  =  Ek + Ep

At different points in the swing, the Energy will be in different forms:-

maximum kinteic energy at lowest point of swing - minimum gravitational potential energy. at the highest point of swing there is no kinetic energy, and maximum gravitational potential energy.

As the Total Energy is conserved:-

Ep (Top)   =  Ek (Bottom) 

m x g x h  =  1/2 x m x v2

2gh = v2

Note - The above calculation shows that Mass has no impact on the speed of a falling object. For more information see  Applications of Forces - Unit 1 

Example 3 - 

What is the maximum velocity of a pendulum if it is raised to a height of 0.2 m above its equilibrium point?

V  =  sqrt (2 x 9.8 x 0.2) 

V  =  1.98 ms-1