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!

A safe held by a string breaks and lands on a spring

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1209 kg safe is 1.54 m above a heavy-duty spring when the rope holding the safe breaks. Thr safe hits the spring and compress it 53 cm. What is he spring constant of the spring?


    2. Relevant equations
    K=mV^2/2
    U=mgh
    Ki+Ui=Kf+Uf


    3. The attempt at a solution

    I thought I understood momentuem and energy but not anymore. Here is what I did....

    Ui=mgh= 2*9.8*1.54
    Kf=mV^2/2

    Then solved for V and got 5.494 m/s

    Then used that velocity for the spring constant by using......

    U=k(delta x)^2/2
    K=mV^2/2

    then solved for k and got 129912 N/m...... but that is coming up wrong what am i doing wrong.
     
  2. jcsd
  3. May 26, 2009 #2
    any ideas?
     
  4. May 26, 2009 #3

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well for one thing you are using m = 2 kg when it is given that m = 1209 kg. Then you neglect the gravitational potential energy in the 2nd set of equations. It is better to do this all in one step: Initial energy = Final energy, where v_initial = v_final = 0.
     
  5. May 26, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi talaroue! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    No, the safe stops moving 53cm lower. :wink:
     
  6. May 26, 2009 #5
    JAY:

    so your saying that i should just use U=mgh and U=k(delta x)^2/2....solve for k and get....

    k=2mgh/delta x^2

    TINY TIM:

    so I didn't go about the problem wrong I just have to add the .53 m to the hieght?
     
    Last edited: May 26, 2009
  7. May 26, 2009 #6

    Astronuc

    User Avatar

    Staff: Mentor

    The safe starts 1.54 m (154 cm) above the spring and then travels another 0.53 m after contact. That must be considered with respect to gravitational potential energy.
     
  8. May 26, 2009 #7

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, but 'h' is as noted by others, and don't forget to correct your value of 'm'.
     
  9. May 26, 2009 #8
    ahhh I see at least I had the right idea just need to be more careful. If you guys don't mind I am having problems with another problem as well.... it might be a same problem i am going to go back and look at it but I made a thread about it as well....

    Second problem I have issues with
     
  10. May 26, 2009 #9
    I wasn't using 2 as my m, that was just noting the second equation is U=k(delta x^2)/2 I didn't combined them when i posted that, i see what you are saying though.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: A safe held by a string breaks and lands on a spring
  1. Landing on bed springs (Replies: 2)

  2. Springs and strings (Replies: 8)

Loading...