1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Loss of potential energy

Tags:
  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-5-22_20-54-49.png

    2. Relevant equations
    GPE=mgh

    3. The attempt at a solution
    I don't understand why the answer is B. My approach is assigning h a random value, say 10m. When the depths of water in both vessels are equal, the height is 5m. Therefore, since the mass of water in both vessels are still m, shouldn't the loss in GPE be half? I don't know why I'm wrong.
     
  2. jcsd
  3. May 22, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    For an extended object, the gravitational potential energy is determined by the location of the center of gravity of the object. Consider the height of the center of gravity before and after.
     
  4. May 22, 2015 #3
    I still don't get one quarter. I will only get that answer if I don't consider the gain in GPE in vessel Y.
     
  5. May 22, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What is the initial height of the center of gravity? The final height?
     
  6. May 22, 2015 #5
    1/2h and 1/4h?
     
  7. May 22, 2015 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. So, what are the initial and final values of gravitational PE?
     
  8. May 22, 2015 #7
    1/2mgh - 1/4mgh?
     
  9. May 22, 2015 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK, that would be the loss of gravitational PE.
     
  10. May 22, 2015 #9
    But how about the gain in GPE in vessel Y?
     
  11. May 22, 2015 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    That is already taken care of. If you consider only vessel X, then the change in PE would be

    PEf - PEi = (m/2)g(h/4) - mg(h/2) = (1/8)mgh - (1/2)mgh = -(3/8)mgh.

    Vessel Y gains PE in the amount (m/2)g(h/4) = (1/8)mgh.

    Overall, there is a change of -(3/8)mgh + (1/8)mgh = -(1/4)mgh.

    But you don't need to consider each vessel individually. Just consider the entire volume of water and work with the center of gravity of the entire volume.
     
  12. May 22, 2015 #11
    Oh, thanks. I didn't know that I should use centre of gravity to solve this question. And what do you mean by an extended object?
     
  13. May 22, 2015 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Extended just means that the mass is spread out over some region rather than being concentrated at a point.
     
  14. May 22, 2015 #13
    Ok, thanks again:oldsmile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Loss of potential energy
  1. Energy loss (Replies: 1)

  2. Loss of energy (Replies: 5)

  3. Energy Loss (Replies: 5)

  4. Loss of Energy (Replies: 0)

Loading...