The aim of this question is the application of the** energy conservation law**Â (**kinetic energy** and **potential energy**).

From the definition of the **energy ****conservation law,** any form of energy can neither be **destroyed nor created**. However, energy may be interconverted between its different forms.

The **kinetic energy** of a body denotes the energy it possesses **due to its motion**. This is mathematically given by the following **formula**:

\[KE \ = \ \dfrac{ 1 }{ 2 } m v^{ 2 } \]

Where $ m $ is the **mass** and $ v $ is the **speed** of the body.

**Potential energy** is the amount of energy a body possesses **due to its position** within an energy field such as a **gravitation field**. The potential energy of a body due to the gravitational field can be calculated using the following **formula**:

\[ PE \ = \ m g h \]

Where $ m $ is the **mass** and $ h $ is the** height of the body**.

## Expert Answer

According to the** law of conservation of energy**:

\[ PE \ = \ KE \]

\[ m g h \ = \ \dfrac{ 1 }{ 2 } m v^{ 2 } \]

\[ g h \ = \ \dfrac{ 1 }{ 2 } v^{ 2 } \]

\[ v^{ 2 } \ = \Â 2 g h \]

\[ v \ = \Â \sqrt{ 2 g h } \ … \ … \ … \ ( 1 ) \]

**Substituting** values:

\[ v \ = \Â \sqrt{ 2 ( 9.8 \ m/s^{ 2 } ) ( 10 \ m ) } \]

\[ v \ = \Â \sqrt{ 196 \ m^{ 2 }/s^{ 2 } } \]

\[ v \ = \Â 14 \ m/s \]

According to the **2nd law of motion**:

\[ F \ = \ m a \]

\[ F \ = \ m \dfrac{ \delta v }{ t }\]

\[ F \ = \ m \dfrac{ v_f \ – \ v_i }{ t } \]

**Since $ v_f = v $ and $ v_i = 0 $:**

\[ F \ = \ m \dfrac{ v \ – \ 0 }{ t } \]

\[ F \ = \ m \dfrac{ v }{ t } \ … \ … \ … \ ( 2 ) \]

\[ F \ = \ ( 70 \ kg ) \dfrac{ ( 14 \ m/s ) }{ ( 1 \ s ) }\]

\[ F \ = \ ( 70 \ kg ) ( 14 \ m/s )\]

\[ F \ = \ 980 \ kg m/s \]

\[ F \ = \ 980 \ N \]

## Numerical Result

\[ F \ = \ 980 \ N \]

## Example

A **60 kg diver** makes a dive and **stops after 1 second** at a **height of 15 m.**Â Calculate the force in this case.

**Recall equation (1):**

\[ v \ = \Â \sqrt{ 2 g h } \]

\[ v \ = \Â \sqrt{ 2 ( 9.8 \ m/s^{ 2 } ) ( 15 \ m ) } \]

\[ v \ = \Â \sqrt{ 294 \ m^{ 2 }/s^{ 2 } } \]

\[ v \ = \Â 17.15 \ m/s \]

**Recall equation (2):**

\[ F \ = \ m \dfrac{ v }{ t } \]

\[ F \ = \ ( 60 \ kg ) \dfrac{ ( 17.15 \ m/s ) }{ ( 1 \ s ) }\]

\[ F \ = \ ( 60 \ kg ) ( 17.15 \ m/s )\]

\[ F \ = \ 1029 \ kg m/s \]

\[ F \ = \ 1029 \ N \]