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Conservation of Energy Problems

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a rock with mass m dropped from rest off a cliff of height h.

    1. If we want to apply conservation of energy to the falling rock, what do we need to include in our system?
    2. Let us set the gravitational potential energy V=0 when the rock is at the bottom of the cliff. With this convention, what is the gravitational potential energy and the kinetic energy of the system when the rock is at the top of the cliff?
    3. What is the gravitational potential energy and kinetic energy of the system just before the rock hits the ground at the bottom of the cliff?
    4. Using your answers to parts 2 and 3 as the end points, make a rough sketch of gravitational potential energy and kinetic energy vs. time during the entire period of the rock's motion from top to bottom of the cliff.

    2. Relevant equations

    KE = 1/2 m v^2
    PE = mgz

    3. The attempt at a solution

    1. earth, rock, and cliff?
    2. PE = m * 9.8 m/s^2 * h; KE = 0?
    3. PE = m * 9.8 m/s^2 * h? KE = (1/2)m(h/t)^2?
    4. Is this like an x-y graph, where x is time and y is energy? if so, then gravitational energy would be like a straight line from top to 0, and KE would be like from the origin to the top right, so like two straight lines criss-crossing; an X
     
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  3. Sep 19, 2007 #2

    learningphysics

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    Number 3 is wrong. how are you getting mgh at the bottom when the gravitational potential energy at the top is mgh?
     
  4. Sep 19, 2007 #3
    well i wrote mgh, because i don't know any information other than the variables..i don't know the height and m and g are constants.
    or, do i know the height? it's not 0 since the question says right before. so what should h be, or how can i calculate it? thanks
     
  5. Sep 19, 2007 #4

    learningphysics

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    the height is 0 at the bottom... gravitational potential energy = mg(0) = 0... also the question says that the gravitational potential energy is taken to be 0 at the bottom.
     
  6. Sep 20, 2007 #5
    yes, the height is 0 at the bottom, but question 3 asks what is the gravitational potential energy just before the rock hits the ground. in other words, it hasn't reached height = 0 yet..
     
  7. Sep 20, 2007 #6

    learningphysics

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    I think they mean for you to take height = 0. The reason they say "before the rock hits the ground" is so that you know that a collision hasn't taken place... because once it hits the ground we have more to deal with than kinetic energy and gravitational potential energy... we have heat, sound etc...
     
  8. Sep 20, 2007 #7
    Alright, what about KE final? Is it KE = (1/2)m(h/t)^2?
    And does 4 look right?

    4. Is this like an x-y graph, where x is time and y is energy? if so, then gravitational energy would be like a straight line from top to 0, and KE would be like from the origin to the top right, so like two straight lines criss-crossing; an X

    Thanks for your help, learningphysics.
     
  9. Sep 20, 2007 #8

    learningphysics

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    No. that's not right. Use conservation of energy. What is the total energy at the top? What is the total energy at the bottom? What must the kinetic energy at the bottom be?

    they won't be straight lines though... try to get the equation for kinetic energy in terms of time.
     
  10. Sep 20, 2007 #9
    ok, i think i got it

    conservation of energy is 0 = change in kinetic energy + change in potential energy

    so (Kf - Ki) + (Vf - Vi)
    since Ki and Vf = 0, we get

    0 = Kf - Vi, which is
    0 = (1/2)mv^2 - mgh
    solving for v, we get
    v = sqrroot(2gh)

    is that right?
    now how do i graph this as an "endpoint" i don't quite get that
     
  11. Sep 20, 2007 #10

    learningphysics

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    Yes. What is the kinetic energy at the bottom before it hits the ground?

    KE = (1/2)mv^2

    What is v in terms of time?
     
  12. Sep 20, 2007 #11
    Okay, so if vf = sqrroot(2gh), then Kf = (1/2)m(2gh) = mgh (????)
    that looks like potential energy...is that right?

    now vf = sqrroot(2gh)..
    there's no t except in "g"
    perhaps i can extract it somehow

    v^2 = 2gh
    [ (1/2)v^2 ] / h = 9.8 m / s^2
    [ [ (1/2)v^2 ] / h ] / 9.8 m = s^2
    sqrroot[ [ (1/2)v^2 ] / h ] / 9.8 m] = s
    well that gives me seconds, which is time, right?

    well, if not straight lines, then curves, right? KE would curve from origin to top right and PE would curve from top left to bottom right 0, right?

    EDIT: also, I added "air" to #1, is that right? since the question doesn't specify whether to consider a frictionless free fall or not...
     
    Last edited: Sep 20, 2007
  13. Sep 20, 2007 #12

    learningphysics

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    yes... but there is a simpler way to look at it. Energy is conserved. The total energy at the top is mgh (gpe = mgh ke = 0). By conserveration of energy, total energy will remain mgh (until a collision or something takes place)... so total energy at the bottom is also mgh. we know that gpe = 0 at the bottom... KE + gpe = mgh, KE + 0 = mgh... so KE = mgh at the bottom.

    you almost did the same there here with:
    At this point you could have continued with:

    Kf = Vi
    Kf = mgh

    and finished there.

    Now, what is v in terms of t... using the knowledge that acceleration is -g? You know that v0=0.
     
  14. Sep 20, 2007 #13
    If energy is conserved, then air does not belong to the system?

    Okay, could you give me some more pointers with "what is v in terms of t"?

    This was my previous attempt:
    v^2 = 2gh
    [ (1/2)v^2 ] / h = 9.8 m / s^2
    [ [ (1/2)v^2 ] / h ] / 9.8 m = s^2
    sqrroot[ [ (1/2)v^2 ] / h ] / 9.8 m] = s

    What did I do wrong? And what is the purpose of solving v in terms of t? For the graph? I'm a little confused. Thanks again.
     
  15. Sep 20, 2007 #14

    learningphysics

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    yes. we'd leave air out of the system... otherwise you have to deal with energy lost to the air etc...

    Well... v is undergoing constant acceleration... so therefore v = v0 - gt = 0 - gt = -gt

    Then substitute into KE = (1/2)mv^2

    so KE = (1/2)m(-gt)^2 = (1/2)mg^2t^2

    Yes, the reason is for the plot... now you can see how KE changes with time.

    Now, you can also get gravitational potential energy at any time t, using the knowledge that total energy is mgh
     
  16. Sep 20, 2007 #15
    Now is this: KE = (1/2)mg^2t^2
    total KE?
    because i already established that KEi = 0 and KEf = mgh; the sum should be mgh, right?
    i don't understand what this KE is
     
  17. Sep 20, 2007 #16

    learningphysics

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    You need to plot KE vs. time, and potential energy vs. time...

    You know KE at the top = 0. so at t = 0, KE = 0. You know KE at the bottom = mgh... but what is KE in between (while it is falling)?
     
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