### Table of Contents

### Definition

**Mechanical energy** is the energy possessed by an object due to its motion or position above ground. In other words, mechanical energy is the sum of kinetic energy and potential energy.

Mechanical energy can be either only kinetic energy (energy due to motion) or only potential energy (due to position) or a combination of kinetic and potential energy.

Examples of object with mechanical energy:

- a moving vehicle has mechanical energy due to its speed (kinetic energy)
- a moving baseball has mechanical energy due to both its high speed (kinetic energy) and its vertical position above the ground (gravitational potential energy)
- a book at rest on the top shelf of a locker has mechanical energy due to its vertical position above the ground (gravitational potential energy)
- a weight lifted high above a weightlifter’s head has mechanical energy due to its vertical position above the ground (gravitational potential energy)
- a drawn (loaded) bow possesses mechanical energy due to its stretched position (elastic potential energy)

Mechanical energy is subject to the **principle of conservation of mechanical energy**, which means that the mechanical energy of an isolated system remains constant in time, as long as the system is free of friction and other non-conservative forces. This means that the potential energy is converted into kinetic energy and kinetic energy is converted into potential energy, in such a way that their sum is always constant.

In a mechanical system like a **swinging pendulum**, subjected to the conservative gravitational force where frictional forces like air drag and friction at the pivot are negligible, energy passes back and forth between kinetic energy and potential energy but never leaves the system.

The pendulum reaches greatest kinetic energy and least potential energy when in the vertical position, because it will have the greatest speed and be nearest the Earth at this point.

On the other hand, it will have its least kinetic energy and greatest potential energy at the extreme positions of its swing, because it has zero speed and is farthest from Earth at these points.

### Formula

Mechanical energy is calculated by adding together the kinetic energy and potential energy. The formula (equation) for mechanical energy is [1]:

_{k}+ E

_{p}

where:

- E [J] – mechanical (total) energy
- E
_{k}[J] – kinetic energy - E
_{p}[J] – potential energy

In most physics books the notation for kinetic energy is K [J], for potential energy is U [J] and for mechanical energy is E [J]. This gives the formula for mechanical energy as:

We know that kinetic energy is calculated as:

^{2}) / 2

And potential energy is calculated as:

If we replace the formulas for kinetic energy and potential energy in equation (1) or (2), we get the formula (equation) for mechanical energy as:

^{2}/ 2 + g · h)

where:

- m [kg] – mass
- v [m/s] – speed
- g [m/s
^{2}] – gravitational acceleration - h [m] – height (measured from the surface of the Earth)

The higher the distance from the surface of the Earth or the higher the speed of the object, the higher the **mechanical energy**. In an isolated system the **mechanical energy** at an initial state E_{i} [J] is equal with the **mechanical energy** at a final state E_{f} [J].

_{i}= E

_{f}

Using equation (2), we can write **principle of conservation of mechanical energy** as:

_{i}+ U

_{i}= K

_{f}+ U

_{f}

The unit of measurement of **mechanical energy** is **joule** [J].

### Example

Calculate the mechanical energy of a 3 kg bowling ball when it’s held at 2 m above ground. Let’s consider that the ball is dropped on the ground. What is the speed of the ball at 70% of the initial height? What about 30%? What is the impact speed of the bowling ball when it touches the ground?

Let’s divide the problem to be solved in four states:

- initial state (i): is the state in which the bowling ball is still held (stationary)
- intermediate state (a): is the state in which the bowling ball is at 70% of the initial height from the ground and its moving towards the ground
- intermediate state (b): is the state in which the bowling ball is at 30% of the initial height from the ground and its moving towards the ground
- final state (f): is the state in which the ball hits the ground

**Step 1 (i)**. Calculate the mechanical energy in the initial state:

_{i}= K

_{i}+ U

_{i}

Since the object is stationary, the kinetic energy will be zero. This means that in the initial state the bowling ball only has potential energy.

_{i}= U

_{i}= m · g · h = 3 · 9.81 · 2 = 58.86 J

**Step 2 (a)**. In the intermediate state (a) the mechanical energy is made up from both kinetic energy and potential energy. The ball is moving thus has a certain speed (kinetic energy) and is also above ground thus has potential energy.

_{a}= K

_{a}+ U

_{a}

First we’ll calculate the potential energy of the ball:

_{a}= m · g · (0.7 · h) = 3 · 9.81 · (0.7 · 2) = 41.202 J

Since the law of conservation of mechanical energy applies, the mechanical energy in state (a) is equal with the mechanical energy in the initial state (i):

_{a}= E

_{i}

Which means that the kinetic energy in state (a) is the difference between the mechanical energy in the initial state and the potential energy in state (a). With other words, if from the total initial mechanical energy we subtract the potential energy in state (a), what remains is the kinetic energy in state (a):

_{a}= E

_{i}– U

_{a}= 58.86 – 41.202 = 17.658 J

Since we know the mass of the object, we can calculate the speed from the kinetic energy formula (3):

\[v_{a} = \sqrt{\frac{2 \cdot K_{a}}{m}} = \sqrt{\frac{2 \cdot 17.658}{3}} = 3.431 \frac{\text{ m}}{\text{ s}}\]**Step 3 (b)**. Similar to intermediate state (a), in the intermediate state (b) the mechanical energy is made up from both kinetic energy and potential energy. The ball is moving thus has a certain speed (kinetic energy) and is also above ground thus has potential energy.

_{b}= K

_{b}+ U

_{b}

First we’ll calculate the potential energy of the ball:

_{b}= m · g · (0.3 · h) = 3 · 9.81 · (0.3 · 2) = 17.658 J

Since the law of conservation of mechanical energy applies, the mechanical energy in state (b) is equal with the mechanical energy in the initial state (i):

_{b}= E

_{i}

Which means that the kinetic energy in state (b) is the difference between the mechanical energy in the initial state and the potential energy in state (b). With other words, if from the total initial mechanical energy we subtract the potential energy in state (b), what remains is the kinetic energy in state (b):

_{b}= E

_{i}– U

_{b}= 58.86 – 17.658 = 41.202 J

Since we know the mass of the object, we can calculate the speed from the kinetic energy formula (3):

\[v_{b} = \sqrt{\frac{2 \cdot K_{b}}{m}} = \sqrt{\frac{2 \cdot 41.202}{3}} = 5.241 \frac{\text{ m}}{\text{ s}}\]**Step 4 (f).** The mechanical energy in the final state (f) is calculated as:

_{f}= K

_{f}+ U

_{f}

In the final state (f), the ball hits the ground thus its potential energy is zero. In this state, the mechanical energy is made up entirely from kinetic energy.

_{f}= K

_{f}= (m · v

_{f}

^{2}) / 2

In the same time, the principle of conservation of mechanical energy applies, which means that the kinetic energy in the final state (f) is equal to the initial potential energy.

_{f}= U

_{i}

From this we can calculate the impact speed of the ball:

\[v_{f} = \sqrt{\frac{2 \cdot K_{f}}{m}} = \sqrt{\frac{2 \cdot 58.86}{3}} = 6.2642 \frac{\text{ m}}{\text{ s}}\]### Calculator

The mechanical energy calculator allows you to calculate the mechanical energy of a body (object) with a given mass, speed and height above the ground. You need to enter the mass, speed and height parameters and choose the desired unit of measurement.

The default unit of measurement for energy is **Joule**. If you want the result displayed in another unit, use the dropt down list to choose and click the CALCULATE button again.

### References

[1] David Halliday, Robert Resnick, Jearl Walker, Fundamentals of Physics, 7th edition, John Wiley & Sons, 2004.

[2] Benjamin Crowell, Light and Matter – Physics, 2007.

[3] Raymond A. Serway and John W. Jr. Jewett, Physics for Scientists and Engineers, 6th edition, Brooks/Cole Publishing Co.,2004

[4] Jiansong Li, Jiyun Zhao, and Xiaochun Zhang, A Novel Energy Recovery System Integrating Flywheel and Flow Regeneration for a Hydraulic Excavator Boom System, Energies 2020.

[5] Leo H. Holthuijsen, Waves in oceanic and coastal waters, Cambridge University Press, 2007.

[6] Kira Grogg, Harvesting the Wind: The Physics of Wind Turbines, Carleton College, 2005.