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Conservation of energy, object at rest

  1. Feb 27, 2015 #1

    Say we have an object released at rest, separated by a certain distance..blah, blah, blah. When we use conservation of energy here, the kinetic energy would have a initial velocity right?
    So, K1+U1=K2+U2 assuming the object eventually comes to a rest...so U1 = 0, K2 = 0, so we would have K1 = U2 right, unfortunately I am unable to find some problems similar to such and would like to know if my thinking is correct, thanks :)!
  2. jcsd
  3. Feb 27, 2015 #2


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    Sounds like you want to determine how high a ball thrown vertically upwards will fly, based on its initial speed. You can use your approach to derive a general formula.
  4. Feb 27, 2015 #3
    Yes, sort of, I was thinking more so for planetary motions, so lets say: If two masses are separated by a radius...and released from rest, I would say the same thing applies right?
  5. Feb 27, 2015 #4


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    Yes, ignoring air resistance, the ball will come back down with the same speed, as it was thrown up.
  6. Feb 28, 2015 #5
    Nope. Velocity(U1) is not a form of energy. Energy associated with motion is kinetic energy. Therefore K1 ≠ U1.

    For calculating the speeds of objects in various contexts, you can use the following kinematic formulas:
    1) at = v-u
    2) S = 0.5at² + ut
    3) 2aS = v²-u²
    Last edited: Feb 28, 2015
  7. Feb 28, 2015 #6


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    This is pretty much all "blah, blah, blah". You have an object, "separated by a certain distance"- from what? Did you mean two objects? Are they attracted to one another by gravity or an electric force? And kinetic energy is NOT an object so does NOT have a velocity!

    It would help if you told us what "K1", "U1", "K2", and "U2" mean. I might guess that "K1" and "K2" are kinetic energy but what are "U1" and "U2"? Because of "conservation of energy" I would think "potential energy" but then there is no reason for potential energy to be 0 when the object is at rest. If you mean U1 and U2 to be speeds, you cannot add energy and speed.

    I think you need to review basic definitions and formulas.
  8. Feb 28, 2015 #7


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    KE doesn't "have a velocity".

    You said the object was released at rest so it's velocity is zero and it's KE is zero.

    You haven't defined what K and U are but never mind.

    Just before they collide the sum of their potential and kinetic energy will be the same as they started with.

    It's more difficult to include what happens after the collision. Some of the energy in the system might be turned into heat or chemical bonds. All you can say is that the sum of all energy in the system is conserved.

    Edit: If you look up inelastic collisions you will see statements that say energy is not conserved. That's because they are only considering the conservation of KE or PE not other forms of energy such as thermal or chemical. If you draw the right boundary around your system and include all forms of energy then energy is always conserved.
  9. Mar 6, 2015 #8
    Hey CW, sorry this question was very vague I agree as I look at it now.

    But, you said KE doesn't have a velocity? Are you just saying that because initially the system was at rest (so KE = 0)? Because KE = 1/2mv^2.
  10. Mar 7, 2015 #9


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    All I meant is that in general Energy doesn't have velocity. It's objects that have velocity and as a result the object has kinetic energy.
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