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Mass-Spring Systems

  1. Dec 10, 2004 #1
    1.) A light horizontal spring has a spring constant of 182 N/m. A 2.93 kg block is pressed against one end of the spring, compressing the spring 0.144 m. After the block is released, the block moves 0.229 m to the right before coming to rest. The acceleration of gravity is 9.81 m/s2.

    a.) What is the coeficient of kinetic friction between the horizontal surface and the block?

    2.) A mass-spring system oscillates with an amplitude of 3.3 cm. The spring constant is 251 N/m and the mass is 0.309 kg.

    a) Calculate the mechanical energy of the mass-spring system. Answer in units of J.

    b) Calculate the maximum acceleration of the
    mass-spring system. Answer in units of m/s2.
  2. jcsd
  3. Dec 10, 2004 #2


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    what have you done?

    1) Application of Newton's 2nd Law and Hooke's Law.
    Alternatively: Change in Mechanical energy = work done by friction

    2) Are you familiar with [itex] E = \frac{1}{2}kA^2 [/itex] ?

    Are you familiar with [itex] a_{max} = \pm A \omega ^{2} [/itex]?

    Are you familiar with [itex] \omega^{2} = \frac{k}{m} [/itex]?
  4. Dec 10, 2004 #3
    question 1 a)

    F_{spring} = kx

    We know k and x, so then we can solve for F_spring.
    F_spring = 26.20800

    W = F_{net}\Delta x = \frac{1}{2}kx^2

    note that Delta x is not the same as x.
    delta x is the displacement of the block after it's released
    and x is the compression. F_net is not the same as the force of the spring because F_net is the sum of all forces which includes the frictional force.

    So, using that equation we can find F_net. This gives F_net = 8.24

    F_{net} = F_{spring} - F_{friction} = F_{spring} - mg \mu

    Remember that the frictional force is the co-efficient times the normal force.
    so now - if u sub in F_net, F_spring, m, and g... you can solve for the coefficient.

    I found that:

    \mu = 0.62547886

    If you find any mistakes, feel free to comment.
  5. Dec 10, 2004 #4
    I have never worked problems with mass-spring systems before. I am familiar with Newton's 2nd Law and Hooke's Law. I just do not know how to incorporate friction in to what I already know. Friction is a completely new concept that I'm having trouble incorporating in to problems.
  6. Dec 10, 2004 #5


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    Just treat friction as another force, except it always opposes motion, and also remember Friction magnitude is given by

    [tex] F_{f} = \mu N [/tex]
  7. Dec 10, 2004 #6
    Yupp, friction is just another force.
    Looking at the Spring force ( hooke's law), F = kx: you can see that both equations has a constant. In Hooke's Law, it's proportional to the compression, which is completely intuitive, if you press harder then there will be a larger force. In Friction, it's proportional to the normal force, that's also intuitive, if the floor is at an angle, then the frictional force will be 0, when the angle is 90 degrees, there is no friction (cos 90 = 0), very intuitive. If the object is heavier than it will have more frictional force, for example pushing a table compared to a refrigerator. If the gravitational force is higher then, of course it will have more frictional force. The normal force is basically the apparent weight.
    Last edited by a moderator: Dec 10, 2004
  8. Dec 10, 2004 #7
    futb0l, regarding problem 1a, the answer is incorrect. I have to submit the work via web, so I automatically know if the answer is correct or not. I am so lost in the book, the book is Cal-based, but the teacher only uses Trig in class.
  9. Dec 10, 2004 #8


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    Well if i'm not mistaken

    The block starts with spring potential energy compressed at a distance x, on a surface with friction, the block is then released amd covers a distance d then comes to a halt, so the spring potential energy converts into kinetic energy which is then lost by friction.

    Ok i believe i made a mistake, it should be

    [tex] \frac{1}{2}kx^2 = \mu mg (d+x) [/tex]
    Last edited: Dec 10, 2004
  10. Dec 10, 2004 #9
    Via the web - that's cool!
    I wish I could do that @ school but that won't happen. :P
  11. Dec 10, 2004 #10
    ok - my final equation is:

    \mu = \frac{kx}{mg}( \frac{1}{2d}x - 1 )

    totally different from urs ...
    Last edited by a moderator: Dec 10, 2004
  12. Dec 10, 2004 #11
    For 2a) - I think the amplitude is the compression.. so it should be just
    * remember to convert cm to m.

    \frac{1}{2}kx^2 = \frac{1}{2}(251)(0.033)^2 = 0.1366695J

    Looks too low - maybe I am wrong.


    F = ma = -kx


    a = -kx/m

    you will find that a = 26.8058252

    hmmm.. this number seems to big, check if this is right..
    give me a few mins to think more about it.
  13. Dec 10, 2004 #12
    Those were all right. Thank you both so much for your help, hopefully now I can do the rest of the problems. THANKS A LOT!!!!
  14. Dec 10, 2004 #13
    No problems - good luck.
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