1. Limited time only! Sign up for a free 30min personal 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!

Velocity and Distance involving springs

  1. Dec 17, 2017 #1
    1. The problem statement, all variables and given/known data
    A spring with an elastic constant of 120 N/m is compressed 0.2 m, then launches a block of mass 0.5 kg on a table with friction 0.49, and it travels 0.9 m before stopping. If the block is then released at the edge of a table top 1.2 m above the ground, how far will the block travel?

    2. Relevant equations
    F = 1/2KX2
    Vf2 - Vi2 = 2a(delta x)
    Vf2 - Vi2 = -2g(delta y)
    3. The attempt at a solution
    I am stumped. I think these equations are relevant, but I'm not totally sure. I already solved for the friction coefficient, so some of those numbers may now be unimportant.
     
  2. jcsd
  3. Dec 17, 2017 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This question is not very clear. Are we to assume that the block is at the end of the same spring that is compressed by the same amount before it is released?
     
  4. Dec 17, 2017 #3
    Yes sorry, the block is launched by the same spring off the table.
     
  5. Dec 17, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Consider the launch velocity.
     
  6. Dec 17, 2017 #5
    I don't know how to find the launch velocity. Do I set the force of the spring equal to kinetic energy?
     
  7. Dec 17, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, a force cannot equal energy, but that's nearly right.
     
  8. Dec 17, 2017 #7
    Well, I have absolutely no clue what I'm doingas finals week has left my mind in shambles. If it's nearly right, what can I do to tweak it and make it correct? I see that F = KE/deltaX, but in this case would the deltaX be the compression x or the x it travels?
     
  9. Dec 17, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is only valid for a constant force acting over distance x. When a spring expands or contracts a distance x the force is not constant.
    Look to the relevant equations you quoted.
     
  10. Dec 17, 2017 #9
    Err, do I find it with the x equation? I have delta x but that's all. I'm afraid I'm not really grasping what you're hinting at.
     
  11. Dec 17, 2017 #10
    Would I set 1/2mv2 equal to 1/2mk2? That's what I just found somewhere else, but I thought you told me that was incorrect (maybe my wording was just poor)
     
  12. Dec 17, 2017 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, set the PE in the spring equal to the KE produced.
    No, you wrote
    That would mean an equation like kx=½mv2.
    As I replied, force is not energy.
     
  13. Dec 17, 2017 #12

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Please note that 1/2mv2 is the correct expression for the KE, but 1/2mk2 is not the correct expression for the PE.
     
  14. Dec 17, 2017 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Thanks for catching that, I had not noticed.
     
  15. Dec 24, 2017 #14
    Im not very sure what the question is asking although it seems that you are saying a spring compressed .2m with k=120 N/m launches a block off a table. The question, I think, is how far the block travels.

    First we need to find the horizontal velocity of the block after the spring has relaxed and transformed all of its potential energy into kinetic energy.
    We can write this as ½m(v*v) = ½k(Δx*Δx) where Δx = displacement of the compressed spring. (½k(Δx*Δx) = potential energy of the spring, PEs)
    v = √k(Δx*Δx)/m
    v = 3.1

    Then, we need to find the time it takes for the block to fall. This can be done by the equation:
    Δy = (vi*t)+½a(t*t)
    In this case Δy is the change in height the object will experience (finale - initial) and we know it will land at a height of 0, so Δy = -1.2. We also know that a=-9.8 because we are finding the time it takes for the mass to fall. The initial velocity is 0 because it doesn't have any initial vertical velocity (velocity is a vector). Finally, this equation can be rearranged as:
    t = √2*(-1.2)/-9.8
    t = .495

    This means that the mass will be in the air for a total of .495s before it hits the ground. During this time, if we neglect air resistance, the mass will have a constant horizontal velocity of 3.1m/s. This means that the distance the mass travels horizontal can be shown as:
    Δx = v(t) +½a(t*t) where a=0 because there is no horizontal acceleration. This means that Δx = v(t) = 1.5m.

    I hope that helped!
     
  16. Dec 24, 2017 #15

    CWatters

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is there a diagram? Where is the edge of the table in relation to the relaxed length of the spring? Is some energy used up against friction before it's launched off the table?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Velocity and Distance involving springs
  1. Spring and distance? (Replies: 3)

  2. Velocity and distance (Replies: 3)

  3. Velocity and distance (Replies: 3)

Loading...