**Conservation of Mechanical Energy**

It is important to know what energy is, but it is more important to understand how it behaves. Considering that energy has different types, it can be transformed from one form to another. Energy all around us continuously change and whatever change they may undergo, they always obey the **Law of Conservation of Energy** which states that

**"Energy cannot be created nor destroyed; ****it may be transformed from one form to another, ****but the total amount remains the same."**

**Example 1.**

Consider a 1.0-kg book lifted to a height of 10.0 m. Using \(g=10.0 \;m/s^2\), the work done on the book is:

\(W=F \cdot d\\ \;\;\;\,=mgh\\ \; \; \; \,=(1\;kg)(10\;m/s^2)(10 m)\\ \;\;\;\,=100 \;J\)

From the ground, the book has 0 total energy. As it is lifted to 10 m high, the work done is transformed into potential energy. As the book is dropped, the velocity increases as the height decreases and at the same time, the potential energy is gradually transformed into kinetic energy. See Picture for illustration.

**Example 2.**

A 55-kg swimmer runs and dives from a diving board to the swimming pool 2 m below the board. If he moves 5 m/s as he leaves the board, solve for:

a. Gravitational potential energy when he leaves the board.

b. His kinetic energy when he leaves the board.

c. His total mechanical energy when he leaves the board.

d. His total mechanical energy just before he enters the water.

Note: The water surface is the reference point.

Given:

\(m=55\;kg\\ h=2\;m\\ v=5\;m/s\\ g=9.8\;m/s^2\)

Solution:

**a.** Realtive to the water surface, the GPE is

\(GPE=mgh\\ \quad \quad\;=(55\;kg)(9.8\;m/s^2)(2\;m)\\ \quad\quad\;=1,078\;J \) ** **

**b. **The kinetic energy can be calculated using the following formula.

**\(KE=\frac12 mv^2\\ \quad\;\;=\frac12(55\;kg)(5\;m/s)^2\\ \quad\;\;=687.5\;J \) **

**c. **The total mechanical energy is the sum of the gravitational potential energy and kinetic energy.

**\(TME=GPE \;+\;KE\\ \quad\quad\;\;= 1,078\;J\;+\;687.5\;J\\ \quad\quad\;\;=1,765.5\;J \)**

**d.** Obeying the Law of Conservation of energy, the total mechanical energy of the swimmer just before he enters the water remains the same, thus **TME= 1, 765.5 J.**

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