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3 years ago

Describe how kinetic and potential energy change as a diver climbs up to a diving board and then dives into the water below.

2 answers:

8 0

A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as PEG=mgΔh

What changes is his Gravitational potential energy due to change of height Δh with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.

Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. 1/2mv2.

While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.

Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.

As such energy transformation equation looks like

Gravitational PE+Elastic PE of springboard→Kinetic energy of air and water+Sound energy of splash+thermal energy

sertanlavr [38]3 years ago

6 0

Do you want to know how both of them is changed by that comment and i will answer

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Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
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Maximum speed you could have and still not hit the deer = 24.07 m/s

Explanation:

Let the maximum speed you could have and still not hit the deer be y.

Your reaction time before stepping on the brakes is 0.50s.

Distance traveled during this time = 0.5y

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Remaining distance to deer = 41 - 0.5y

The maximum deceleration of your car is 10 m/s²

We have equation of motion, v² = u² + 2as

Initial velocity, u = y m/s

Final velocity, v = 0 m/s

Acceleration, a = -10 m/s²

Displacement, s = 41 - 0.5y

Substituting,

v² = u² + 2as

0² = y² + 2 x -10 x (41 - 0.5y)

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