Energy is the ability to do work. There are two types of energy that constitute mechanical energy, they are **Potential** energy and **Kinetic **energy.

**Mechanical energy** is the sum of the kinetic energy and potential energy of an object that is used to do a particular work. In other words, it describes the energy of an object because of its motion, position, or both.

Potential energy, P.E = mgh (Energy of a body by virtue of its position or rest or stored energy due to height).

m = mass of the body

g = acceleration due to gravity

h = height to which the object is raised

Kinetic energy, K.E = \( \frac{1}{2} \scriptsize mV^2 \) (Energy of body in motion)

m = mass of the body

V = Velocity of the body

The principle of the conservation of mechanical energy states that the total mechanical energy in a system (i.e., the sum of the potential plus kinetic energies) can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature.

The **law of conservation of mechanical energy** shows that P.E + K.E = constant.

### Machines:

A machine is a device or a contrivance with which force is applied at one end (Effort) to overcome load (L) at the other end.

A small force (effort) can be applied to overcome a larger load.

Examples of such machines are levers, pulleys, inclined planes, screws, hydraulic presses, etc.

### Terms Associated with Machines:

Let’s define some terms that apply to the workings of machines.

### Mechanical Advantage of Force Ratio:

This is the ratio of load to effort. It can be defined as the ability of a machine to overcome a large load through a small effort.

M.A = \( \frac{Load}{Effort} \\= \frac{L}{E} \)

It has no unit and is also referred to as the **Force Ratio**.

Since load is the output force and effort is the input force, Mechanical Advantage can also be defined as:

M.A = \( \frac{Output\:force}{Input\:force} \)

### Velocity Ratio:

This is the ratio of distance moved by effort to the distance moved by load.

V.R = \( \frac{distance \: moved \: by \: effort}{distance \: moved \: by \: load} =\frac{x}{y} \)

When load is greater than effort, M.A > 1, V.R > 1.

### Efficiency:

This is the ratio of work output to the work input of a machine expressed as a percentage.

Efficiency = \( \frac{Work \:output}{Work\: input} \scriptsize \: \times \: 100 \%\)

But work = \( \scriptsize Force \: \times \: distance\: / \: displacement \)

E = \( \frac{Load \: \times \: distance \: moved \: by \: load}{Effort \: \times \: distance \: moved \: by \: effort} \scriptsize \: \times \: 100 \% \)

E = \( \frac{L}{E} \: \times \: \frac{y}{x} \: \times \: \scriptsize 100 \% \)

E = \( \frac{\normalsize\frac{L}{E}}{ \normalsize\frac{x}{y}}\scriptsize \: \times \: 100 \% \)

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

For an ideal machine, the efficiency is 100%, but due to friction, mechanical advantage (M.A) decreases, and this also affects (reduces) the efficiency of the machine.

### Example 1.1.1:

A machine has an efficiency of 60%. If the machine is required to overcome a load of 30 N with a force of 20 N, Calculate its Mechanical Advantage.

**Solution**

Load = 30 N

Effort = 20 N

**Use the formula:**

Mechanical Advantage, M.A = \( \frac{Load}{Effort} \)

M.A = \( \frac{30\:N}{20\: N}\\ = \scriptsize 1.5 \)

### Example 1.1.2:

The velocity ratio of a machine is 5 and its efficiency is 75%. What effort would be needed to lift a load of 150N with the machine?

**Solution**

Velocity Ratio (V.R) = 5

Efficiency = 75% or \( \frac{75}{100} \)

Load = 150 N

First, let’s calculate the Mechanical Advantage (M.A):

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

â‡’ \(\frac{75}{100} = \frac{M.A}{5} \)

M.A = \(\frac{75\:\times \:5} {100} \)

M.A = 3.75

Since we know the load and the mechanical advantage, we can use the formula below to calculate the effort.

M.A = \( \frac{Load}{Effort} \)

3.75 = \( \frac{150}{Effort} \)

Effort = \( \frac{150}{3.75} \)

Effort = 40 N

### Example 1.1.3:

A machine with velocity ratio of 5 requires 1200 J of work to lift a load of 400 N through a vertical distance of 2.0 m. Calculate the efficiency and mechanical advantage of the machine.

**Solution**

Efficiency = \( \frac{Work \:output}{Work\: input} \scriptsize \: \times \: 100 \%\)

Work input = 1200 J

Work output = \( \scriptsize Force \: \times \: distance \)

âˆ´ Efficiency = \( \frac {400 \: \times \: 2}{1200} \scriptsize \: \times \: 100\% \\ = \frac {800}{1200} \scriptsize \: \times \: 100\% \\ \scriptsize = 66.7 \%\)

Since we now have a value for efficiency, we can use the formula below to calculate mechanical advantage.

E = \( \frac{M.A}{V.R}\scriptsize \: \times \: 100 \% \)

\( \frac {66.7}{100} = \frac {M.A}{5} \)M.A = \( \frac {66.7\: \times \:5}{100} \)

= 3.33.

Awesome!

Wow!