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A 1.0 kg rocket dropped from a height of 6.0 m at what height is the rock's genetic energy twice is potential energy
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For the person, who calls kinetic energy as "genetic energy" and misses "rocket" with "rock",
I would NEVER write any solution. He (or she) does not deserve it . . .
But I think about other people and next generations students, who will read my post in the future.
My post is for them.
I will start re-formulating the problem as it should sound.
A 1.0 kg rock dropped from a height of 6.0 m.
At what height is the rock's kinetic energy twice its potential energy ?
In the figure on the right I show a vertical trajectory of the falling rock. * 6 meters
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The intial height is 6 m above the ground level. The current position of the falling rock Y
is "z" meters above the ground level. |
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- z
In this current position, the kinetic energy is twice the body's potential energy. |
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We should find this current coordinate "z". ------- ground level
Due to the Mechanical Energy Conservation law, the kinetic energy of the rock in its current
position "z" is equal to the loss of its potential energy from the height of 6 m to the current level "z"
kinetic energy (z) = loss of the potential energy = mg*(6-z) joules.
The potential energy of the rock at the level "z" is mgz joules.
So, the equation to find z is
mg*(6-z) = 2mgz,
where m is the mass of the body and g is the gravity acceleration.
Canceling the common factor mg in both sides of the last equation, we get
6 - z = 2z.
We simplify this equation
6 = 2z + z, or 6 = 3z,
and find finally
z = 2 meters.
ANSWER. At the height of 2 meters above the ground level, the kinetic energy of the rock is twice its potential energy.
Solved.
