1. Not finding help here? Sign up for a free 30min 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!

Conservation of energy

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle P of mass m is attached to one end of a light string of natural length l whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length 5l/3. Show that if P is projected vertically downwards from A with speed √(3gl/2). P will come to instantaneous rest after moving a distance of 10l/3

    2. Relevant equations

    E.P.E= λx^2/2l, v^2=u^2+2gh, P.E= mgh

    3. The attempt at a solution
    Here, I have been asked to show that the total distance traveled by the particle is 10l/3 which we know is true if we think logically about it because when I take the particle up at A it travels 5l/3 and when it comes back to it's equilibrium position it again traverses 5l/3. When we lift the particle the work done in this case is P.E=mgh and when we project the particle vertically downwards then the work converts P.E into K.E and E.P.E where the equation of velocity is v^2=u^2-2gh → 0=u^2-2gh→ u^2=2gh. Should the total equation for the conservation of energy look like this that describes both the particle's action of going up and coming down
    P.E= E.P.E + K.E
     
  2. jcsd
  3. Mar 1, 2014 #2
    I do not understand you explanation. The particle is not "taken up at A" and it does not travel 5l/3. The problem says that the entire length of the spring is 5l/3 in equilibrium, which a different statement.

    Anyway, all you need here is to compute the total mechanical energy initially and the total mechanical energy at max elongation.
     
  4. Mar 1, 2014 #3
    Shouldn't the equation of conservation of mechanical energy look like this P.E= E.P.E + K.E
    mgh=λx^2/2l + mv^2/2
     
  5. Mar 6, 2014 #4
    When I equate the whole thing like mgh=λx^2/2l + mv^2/2 taking velocity=√(3gl/2) I don't get h= 10l/3 why???????
     
  6. Mar 6, 2014 #5
    I am not sure what λx^2/2l means. Is it (λx^2)/(2l) or is it (λx^2/2)l? And what is λ anyway? This does not look like stiffness in Hooke's law.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted