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!

Work/Engery Problem: Emergency

  1. Nov 1, 2006 #1
    I am trying to do this problem but I have no idea howa to do it. Can someone please give me a step by step explanation of how to do it? Please, thanks

    [​IMG]
     
  2. jcsd
  3. Nov 1, 2006 #2

    rsk

    User Avatar

    hang on, rethinking

    On second thoughts I'd say use energy.

    How much GPE does it lose sliding doen the slope?

    If this is converted to EPE of the spring, what is the extension?


    (First method, using force, gives me an answer eactly half of this .... trying to work out why)
     
    Last edited: Nov 1, 2006
  4. Nov 1, 2006 #3
    I don't get it. I am still confused. Someone please give me a clear explanation. Thanks
     
  5. Nov 1, 2006 #4

    radou

    User Avatar
    Homework Helper

    Or, you can use an energy approach, since I assume that was the topic of your last unit. Did you learn something about conservation of energy?
     
  6. Nov 1, 2006 #5
    Yes, this homework relateds to the Work, Engery stuff. I know that the potetnial engery is mgh, and kenetic engery is 1/2mv^2. Also, for springs, teh engery is 1/2kx^2. But I don't undertsand how to eo this problem
     
  7. Nov 1, 2006 #6

    rsk

    User Avatar

    How much GPE does it lose sliding doen the slope?

    If this is converted to EPE of the spring, what is the extension?
     
  8. Nov 1, 2006 #7
    Well before it falls, the GPE is mgh, where h is xsinθ

    but still i dont' know hwo to slove thep roblem
     
  9. Nov 1, 2006 #8

    rsk

    User Avatar

    No - just look at the change in GPE.

    What distance (vertical) does it move? This allows you to find the change in GPE, which must be converted to EPE
     
  10. Nov 1, 2006 #9
    If you make the ground level to be the stoping point of the block, then we can say that

    sinθ = h / x

    ....wait, nvm, i am confused, can you please give me a detalted explanation
     
  11. Nov 1, 2006 #10

    radou

    User Avatar
    Homework Helper

    Define your system - it consists of a spring and a block. The energy of the system is defined as the sum of kinetic and potential energy, for each part of the system. Energy is conserved. Take two points, 1 and 2. Let point 1 be the point when the block is held, and it posesses gravitational potential energy only. Let point 2 be the point when the block has sled down, and stopped. At that point it posesses no kind of energy, but the spring posesses elastic potential energy.

    Since energy is conserved, the energy at point 1 must equal the energy at point 2. What equation does that give you?
     
    Last edited: Nov 1, 2006
  12. Nov 1, 2006 #11
    So, at point 1, where the block is held initially, I got the following engery equation

    Point 1

    (1/2)mv^2 + mgh = (1/2)kx^2
    0 + mgh = 0
    mgh = 0

    0 for kenetic energy because the block is not moving, and 0 for spring because it is not yet stretched.

    Point 2

    (1/2)mv^2 + mgh = (1/2)kx^2
    0 + 0 = (1/2)kx^2

    0 for potential enegery because it has reached OUR made up groud level
    0 for kenetic enegry because we it stops at this pont.

    So, what do I do now?
     
  13. Nov 1, 2006 #12

    rsk

    User Avatar

    Why is mgh at point 1 zero? It's above your 'ground' level

    What vertical distance h does the block move? (use x and trigonometry)

    Now, use that number to find the change in GPE=mgh

    Now put that = EPE in the spring.
     
  14. Nov 1, 2006 #13

    radou

    User Avatar
    Homework Helper

    Please ignore what I said about the displacement down the incline not being equal to the displacement in the spring. It was pretty stupid, and I deleted it.

    [:smile:]
     
  15. Nov 1, 2006 #14
    Oh okay, how do I do 27?
     
  16. Nov 1, 2006 #15

    rsk

    User Avatar

    I don't understand that. Surely at its lowest point it's come to rest?
     
  17. Nov 1, 2006 #16
    Thats what I am thinking if they are saying the lowest point. But I don'twant to risk putting in the answer as 0 because i am pretty sure they are asking something else.

    What is they are asking you for the acceleartion right before it stop? How do I find the acceleartion fight before it stops?
     
  18. Nov 1, 2006 #17

    radou

    User Avatar
    Homework Helper

    Ok, once again:

    mg x sin(30) = 1/2 k x^2.

    The term on the left side is the energy of the system at point 1, and the term on the right side is the energy of the system at point 2. I hope this is clear now.
     
  19. Nov 1, 2006 #18
    Yea, i figured that aout, but now I am trying to figure out how to do the next question
     
  20. Nov 1, 2006 #19

    radou

    User Avatar
    Homework Helper

    Try using Newton's 2nd law.
     
  21. Nov 1, 2006 #20

    can you please give me the equation? Please? I only have 20 mins to submit thsi answer. please, thanks after the homework, i will pick up the solutions and see how they got the answer. please
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work/Engery Problem: Emergency
  1. Engery Problem (Replies: 3)

Loading...