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A high diver of mass 70.0 kg jumps off a board 10 m above the water. If, 1.0 s after entering the water his downward motion is stopped, what average upward force did the water exert?

The aim of this question is the application of the energy conservation law (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:

\[ PE \ = \ KE \]

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

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

\[ v^{ 2 } \ = \  2 g h \]

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

Substituting values:

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

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

\[ v \ = \  14 \ m/s \]

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. Calculate the force in this case.

Recall equation (1):

\[ v \ = \  \sqrt{ 2 g h } \]

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

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

\[ v \ = \  17.15 \ m/s \]

Recall equation (2):

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

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

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

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

\[ F \ = \ 1029 \ N \]

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