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!

Compression of a spring with friction

  1. Sep 5, 2012 #1
    Hi again

    I try to solve this by my own with little help from the net, but still I'm not sure if it's the correct!

    Here is my work, and please someone tell me if I'm wrong, again. (x = 20,8 m ???)

    Thanks

    P.S. Here is the full problem http://i.minus.com/j8Fbhtd9CJtCQ.jpg [Broken]
     

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 5, 2012 #2
    Some parts of the problem text are washed out, and the solution is out of focus. I don't think you want any guesswork on our part, so you should give us something more readable if you need help.
     
  4. Sep 5, 2012 #3
    Here, more readable:

    A sledge of mass m=2 kg starts with initial velocity Va=7 m/s at ha=40m. Compute, neglecting friction.
    a) the speed Vb at hb=20m

    The sledge then travels from C (hc=30m) to D horizontally on a rough surface with uk=0.6.
    From D the surface becomes frictionless. In E a spring in equilibrium condition (i.e. neither streched nor compressed) is positioned.
    b) Determine the maximum compression of the spring, knowing that CD=20m and the spring constant k=0.2 N/cm.
     
  5. Sep 5, 2012 #4
    PE+KE=Wf+PEs

    mgh + 1/2 mv2= mgμd + 1/2 kx2
    Check for zero PE reference level.
    Gravitational force is conservative, only levels are relevant not the path between the levels.
     
  6. Sep 5, 2012 #5
    So, I take Vc and d as CD?
     
  7. Sep 5, 2012 #6
    So, for v I take Vc and for d = CD?
     
  8. Sep 5, 2012 #7

    CAF123

    User Avatar
    Gold Member

    Yes.
     
  9. Sep 5, 2012 #8
    And then, the final result is x=7.6m if I'm not wrong again?
     
  10. Sep 5, 2012 #9
    Can you show your working.
    Check the unit of spring constant k.
     
  11. Sep 5, 2012 #10

    CAF123

    User Avatar
    Gold Member

    I get a very similar answer(≈7.7m).
     
  12. Sep 5, 2012 #11
    mgh + 0.5mv2=mgμd+0.5ky2
    2x9.8x10 +0.5x2x7x7=2x9.8x0.6x20 + 0.5x0.2/0.01 x y2
    196+49 =235.2 +10xy2
    9.8=10xy2

    y2= 0.98
    y=1m
     
  13. Sep 5, 2012 #12

    CAF123

    User Avatar
    Gold Member

    Why is h = 10 m and v = 7m/s? I computed the velocity at C to be 15.7m/s and the motion is at a height of 30m. (I took the reference of zero potential at ground level).
     
  14. Sep 5, 2012 #13
    As i mentioned on previous reply, it is a conservative force. It is not he path but the different of height that determines the ΔPE. So the ΔPE is height between A and C.
    Conservation of energy. The kinetic energy remain the same. As in projectile, the horizontal velocity remains constant.

    From point A to C no energy dissipated.

    add: If level A equal to level C, do you need to take zero reference at ground level?
     
  15. Sep 5, 2012 #14

    CAF123

    User Avatar
    Gold Member

    Ah of course! But I still don't see why the velocity at C is 7m/s. Can you show this mathematically? I have tried but get values greater than this.
     
  16. Sep 5, 2012 #15

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are too many steps here that cause errors. Eliminate all the intermediate steps. Start at A and end at E. [itex]
    (KE + PE)_{initial- at- A} + W_f = (KE + PE)_{final- at -E}[/itex]
    Work done by friction is numerically a negative value.
     
  17. Sep 5, 2012 #16
    So, guys, what is the final and also the right result?
     
  18. Sep 5, 2012 #17

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh heck you have a good understanding of work energy concepts, so it's just math now for the 2nd part and using g = 10 , then

    1/2(2)(7)^2 + (2)(10)(40) + 0 - .6(2)(10)(20) = 0 + 2(10)(30) + 1/2(20)(x^2)

    From which x = 1 m.

    Which looks like what azizlwl got using a slightly different approach between a different set of points. You should confirm this on your own, as there are lots of ways to skin a cat, as they say.

    Oh sorry I didn't pay attention to part A I guess that's solved already.
     
  19. Sep 5, 2012 #18
    Okay, the part a) is just fine, no problem at all with it...
    Many thanks to everybody, I hope the answer of 1m is correct :P
     
  20. Sep 5, 2012 #19

    CAF123

    User Avatar
    Gold Member

    Can anyone enlighten me as to why the velocity at C is 7m/s? Many thanks
     
  21. Sep 5, 2012 #20

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You should confirm it is correct on your own.....I didn't take the time pouring through your pages of calcs to see why you arrived at a different answer...maybe too many steps and a number got transposed somewhere... or maybe our answer is wrong you know.....:wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook