Kinetic and Potential Energy Relationship Explained - Solve for One-Third KE

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SUMMARY

The discussion focuses on the relationship between kinetic energy (KE) and potential energy (PE) as described by the conservation of mechanical energy. The equation Ei = Ef illustrates that the initial kinetic and potential energies equal the final kinetic and potential energies. A specific problem is solved where a ball dropped from a height of 60 meters reaches a height of 40 meters when one-third of its total energy is kinetic energy. The calculations demonstrate that the total energy at the initial height is 588m, with 196m allocated to kinetic energy at the specified height.

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rkslperez04
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I was in class today and we flew through the relantionship with Kinetic engery and Potential engery rather quickly.

I looked it up in the book but it went over my head.

Ei = Ef

Meaning
Ei = Ef
^ ^
Kinetic intital + Potential intital = Kinetic Final + Potential Final


This relates to the following question:

A ball is dropped from a hieght of 60meters. How high above the ground will it be when one-third of its total engery is KE?


Can you tell me the the Ei = Ef means and then give me a starting point to work the problem?
 
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Kinetic intital + Potential intital = Kinetic Final + Potential Final

That is correct. This an example of conservation of energy, and is known as conservation of mechanical energy (Kinetic and potential). It says, the mechanical energy at any time is a constant, in the absence of forces such as friction.
 
i think
the energy is initial potential energy at 60m and kinetic energy is zero
and as the ball is dropped
it comes down decreesing the potential energy and increasing the kinetic energy and at the utmost bottom, there is maximum kinetic energy
so kinetic energy increases coming downward and potential energy
at 60 meter, total Energy= PE= m*g*h= 9.8*60*h=588m
as illustrated KE= total energy/3 = 588m/3 = 196m
now placing in equation
TE= KE + PE or PE1+KE1 = PE2+KE2,
where KE1=0, PE1=588m, KE2= 196m and PE2=?
0+588m=196m+mgh2 suppose reuired height is h2
cutting m from all equation
588=196+gh2 => gh2=588-196 => h2= 392/g = 392/9.8 = 40
so the height will be 40 meters
 
ohh ok... let me rework the problem now...

soo when you drop a ball... the KE on the ball is 0 being its like Intial Velocity.. no movement yet... but as it goes... it gains kinetic engery ??
 
yup u got it. coz initially the ball is released instead of giving it a force for it to move.as it moves,the ball lose PE while gaining KE meaning energy is conserved. there is another point i want to add...when u throw a ball vertically upwards,initial velocity is not zero...
 

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