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 Potential Energy?

  1. Oct 28, 2006 #1
    A block of mass m = 3.6 kg is dropped from height h = 73 cm onto a spring of spring constant k = 1470 N/m . Find the maximum distance the spring is compressed.

    I tried calculating the gravitational potential energy (mgy) then setting it equal to 1/2kx^2. but its not working out, can anyone help me out?
     
  2. jcsd
  3. Oct 28, 2006 #2

    radou

    User Avatar
    Homework Helper

    It should work out. Did you convert the height h into meters?
     
  4. Oct 28, 2006 #3

    rsk

    User Avatar

    Is that because when the spring is compressed the mass falls a little bit further and loses more gpe?

    So the change in position of the block is 73cm + x, not just 73cm?
     
    Last edited: Oct 28, 2006
  5. Oct 28, 2006 #4
    i converted it into meters, and it doesnt work out still. i dont know whats wrong. actually someone else posted this same question earlier and tried the same method and also got it wrong
     
  6. Oct 28, 2006 #5

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    You have the right idea, this should work.

    Did you use meters? (not cm)
    Did you include the distance x in addition to the height the mass fell?
     
  7. Oct 28, 2006 #6
    no i did not include .73m in addition to x, I am a bit confused by this can someone explain it a little more to me?
     
  8. Oct 28, 2006 #7

    rsk

    User Avatar

    The block falls 73cm and hits the top of the spring.

    The spring compresses. Does the block stay at the height where the top of the spring was[/], suspended in free air? No, it falls a little bit further, a distance equivalent to the compression of the spring.

    So mgh is mg(0.73 + x)

    Equate this to you energy stored in the spring and you get a quadratic to solve for x.
     
  9. Oct 28, 2006 #8
    ok im trying to solve the quadratic right now, think i made a mistake though i got a really large number
     
  10. Oct 28, 2006 #9

    rsk

    User Avatar

    What is your quadratic? And do you know what the answer should be?
     
  11. Oct 28, 2006 #10
    735x^2 - 35.28x - 25.7544 = 0
    not working im getting a very large incorrect number
     
  12. Oct 28, 2006 #11
    i dont know what the answer should be i have to input answers online, and so far they have been incorrect
     
  13. Oct 28, 2006 #12

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    How large? The positive solution to this equation is about 0.22m. The equation is correct.
     
  14. Oct 29, 2006 #13
    yes, i was making very foolish mistakes figured it out eventually. But now I am having problems with this question:

    Tarzan, who weighs 629 N, swings from a cliff at the end of a convenient vine that is 21 m long. From the top of the cliff to the bottom of the swing, he descends by 4.9 m. The vine will break if the force on it exceeds 1430 N. What would the greatest force on the vine be during the swing?

    I am thinking i should set the initial gravitational potential energy equal to the kinetic energy at the bottom of the swing, calculate for v find centripetal accelaeration and use that to calculate the force on the vine. However, I do not know how I can calculate the value of the initial potential energy if I do not know the height of the cliff. Can someone explain that part to me please?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?