Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Spring/distance ball problem

  1. May 22, 2008 #1

    krisrai

    User Avatar

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

    A 3-kg mass starts at rest and slides a distance d down a smooth 30o incline, where it contacts an unstressed spring. It slides an additional 0.2m as it is brought momentarily to rest by compressing the spring (k=400 N/m).

    Find the initial separation d between the mass and the spring.


    2. Relevant equations
    Work=F*d
    Ek=1/2mv^2
    Espring=1/2kx^2
    v2^2-v1^2=2ad
    d=vt
    Fg=mg

    3. The attempt at a solution

    I know that vf=0
    and that v1=0
    I really need help with v2 but im not too sure how to go at it.
    is this way of finding v correct even though the balls velocity is at an angle:
    1/2mv^2=1/2kx^2
    v= sqrt16/3 m/s


    can anybody please help?
     
    krisrai, May 22, 2008
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. May 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I assume v2 is the speed when the ball first contacts the spring. Your calculation is almost right, but you forgot about gravitational PE, which changes as the spring is compressed.

    While there's nothing wrong with calculating v2 to solve this problem, you can also solve this problem without finding v2. Just compare initial and final energy.
     
    Doc Al, May 22, 2008
  4. May 28, 2008 #3

    hi88

    User Avatar

    would someone further clarify how to solve this problem?

    if you are to compare initial and final energy, should it be:
    mgh= 1/2kx^2 ... solve for h?
    h= xsin30
     
    hi88, May 28, 2008
  5. May 28, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Assuming h is measured from the lowest point, yes.
    No. (x is given, by the way.)
     
    Doc Al, May 28, 2008
  6. May 28, 2008 #5

    hi88

    User Avatar

    okay, so here it goes:

    mgh=1/2kx^2
    (3kg)(9.8)(sin30h)= 1/2(400N/m)(0.2m)
    solve for h
    h= 2.72 m

    correct?


    gravitational potential energy being converted to kinetic energy
     
    hi88, May 28, 2008
  7. May 28, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Here, h is the vertical distance between the starting point and the lowest point.
    I think you mean "d" instead of "h". Two problems:
    (1) You forgot to square the 0.2m.
    (2) Here, "d" would be the entire distance the mass travels along the incline, not just the initial separation between the mass and the spring (which is what you need to find).
    More like gravitational PE converted to spring PE.
     
    Doc Al, May 28, 2008
  8. May 28, 2008 #7

    hi88

    User Avatar

    correct, d would be the entire distance the mass travels along the incline.. so then, how would you go about finding the initial separation between the mass and the spring?

    (3kg)(9.8)(sing30d)=0.5(400n/m)(0.2)^2
    d= 0.544 m

    would you subtract 0.2 m?
     
    hi88, May 28, 2008
  9. May 28, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Yes.
     
    Doc Al, May 28, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook