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inkyvoyd

  • one year ago

A m kg block slides along a horizontal surface with friction. The block has a speed v when it strikes a massless spring head-on (as in the figure).

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  1. inkyvoyd
    • one year ago
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    |dw:1444253632076:dw|

  2. inkyvoyd
    • one year ago
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    What minimum value of the coefficient of static friction, µs, will assure that the spring remains compressed at the maximum compressed position? Assume the spring has a spring constant k.

  3. inkyvoyd
    • one year ago
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    I did the system 1/2 mv^2=1/2kx^2+u m g x for energy conservation

  4. inkyvoyd
    • one year ago
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    u m g=k x for spring remaining compressed at the maximum compressed position

  5. Empty
    • one year ago
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    Cool this looks fun, but the thing is we can't use energy conservation here because friction isn't a conservative force, so it's path dependent and not a state function.

  6. inkyvoyd
    • one year ago
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    fudge

  7. Empty
    • one year ago
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    Well let's try to piece this together, it looks like we can lay out a few variables and stuff and see what happens. Give me a sec to fiddle with some symbols on my paper here.

  8. inkyvoyd
    • one year ago
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    the first part of the problem specified the static friction, and asked what the compression would be, and I got that right using \(\large 1/2 mv^2=\int^x_0kx+umg \text{ }ds\) which is how I got \(1/2mv^2=1/2kx^2+umgx\) then I solved for x and put that in, and it said it was right... did I get it right cause I was lucky even though the method is invalid, or did that integral exhibit path dependency?

  9. Empty
    • one year ago
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    Yeah this is fine cause you are only going to the maximum, now if it were to go past that and say go back to spring equilibrium, then you'd feel the friction force again on the way back. So this integral is valid yeah.

  10. Empty
    • one year ago
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    So wait, is u your coefficient of static friction?

  11. inkyvoyd
    • one year ago
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    yeah

  12. Empty
    • one year ago
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    I think we might be able to take the derivative of u with respect to x and set it equal to 0 to find the minimum of u?

  13. inkyvoyd
    • one year ago
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    (sorry for delayed responses, I'm working on other parts of this assignment that aren't so conceptual and more tedious)

  14. Empty
    • one year ago
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    Yeah sure I'm just kinda having fun here anyways I haven't really touched this sorta stuff in a while. I'm not sure about this derivative though, I guess because v and x will be a constant at the maximum.

  15. inkyvoyd
    • one year ago
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    well I got du/dx=-(kx+umg)/(mgx)

  16. inkyvoyd
    • one year ago
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    which, as expected, gives me kx=umg hahaha

  17. inkyvoyd
    • one year ago
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    then I did the system; 1/2 mv^2=1/2kx^2+u m g x kx=umg and got u=(v sqrt(k))/(g sqrt(3m))

  18. inkyvoyd
    • one year ago
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    which... is wrong

  19. inkyvoyd
    • one year ago
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    oh wait that's stupid, that assumes u_s=u_k which is totally not true I'm stoopid

  20. Astrophysics
    • one year ago
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    That should work

  21. Astrophysics
    • one year ago
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    \[kx= u mg\]

  22. inkyvoyd
    • one year ago
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    yeah I got confused cause I assumed u_s=u_k for some reason and then subsituted that equation into the energy equation.

  23. Astrophysics
    • one year ago
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    Ah ok I see

  24. inkyvoyd
    • one year ago
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    figured it out now tho

  25. inkyvoyd
    • one year ago
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    thanks

  26. Astrophysics
    • one year ago
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    Cool question though, thanks for sharing :)

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