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PLEASE help me out coefficient of friction problem

  1. Oct 8, 2004 #1
    PLEASE help me out!!! coefficient of friction problem

    Okay, I've never done this before and I sure hope there is somebody out there who can help me. I've tried answering this question EVERY way possible and cannot come up with the right answer.

    In a circus performance, a monkey is strapped to a sled nd both are given an initial speed of 4 m/s up a 20º inclinedj track. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?

    FYI: Book answer is 1.5 m
     
  2. jcsd
  3. Oct 8, 2004 #2

    Pyrrhus

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    First do the forces analysis for your body [monkey+sled].

    The forces acting are the weight of your body, the normal force of the incline, and the friction force.

    Use newton's 2nd Law

    [tex] \sum_{i}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    Also remember

    [tex] F_{f} = \mu N [/tex]
     
  4. Oct 9, 2004 #3
    Thanks for the help!! And I hate to say it... But I'm still confused. I know the equations, I've tried them all out, and I've drawn free body diagrams... If you can help me out anymore, please do (test on Monday).
     
  5. Oct 9, 2004 #4

    Pyrrhus

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    Show your equations.
     
  6. Oct 9, 2004 #5
    Let's see... I've tried conservation of energy, nonconservation, nonisolated systems, velocity of base equals the square of 2gh, -f(subk)d- 1/2mvf2-1/2mvi2.... umm, -u(subk)mgd=/1/2mvb2.... The only thing I have figured out is that y=.816, Vi=4ms/, Vf=0m/s, and m1+m2= 20kg.. and the kinetic coefficiet is .20. i'm stuck and really frustrated.
     
  7. Oct 9, 2004 #6

    Pyrrhus

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    Ok i did your problem and got the book answer, so the book is right.

    Now let me analyze your work.
     
  8. Oct 9, 2004 #7

    Pyrrhus

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    Why would you use conservation of energy??

    This is definetly NOT a conservative system.

    Instead the Mechanical energy change will be equal to the work done by friction.

    Apply this.

    [tex] \Delta E = W_{f} [/tex]
     
  9. Oct 9, 2004 #8
    Oh!!! If you can show me how to do this problem I'd be so grateful!!! I'll actually get sleep tonite!.... even though I have 10 more problems to go.
     
  10. Oct 9, 2004 #9

    Pyrrhus

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    Ok let's try this

    [tex] E - E_{o} = W_{f} [/tex]

    What is the initial mechanical energy?

    What is the final mechanical energy?

    what is the force of friction equal to?
     
  11. Oct 9, 2004 #10

    Pyrrhus

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    doxi, if you do not understand or cannot answer the questions, say it, i'll help.
     
  12. Oct 9, 2004 #11
    thanks for your help, but nevermind. i give up, i can't figure it out.
     
  13. Oct 9, 2004 #12

    Pyrrhus

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    Ok, i will do it, don't go.
     
  14. Oct 9, 2004 #13

    Pyrrhus

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    Applying:

    [tex] \Delta E = W_{f} [/tex]

    [tex] E - E_{o} = \vec{F_{f}} \cdot \vec{r} [/tex]

    [tex] E - E_{o} = |\vec{F_{f}}|\vec{r}|cos \theta [/tex]

    where [itex] \theta [/itex] is the angle between both vectors, in this case 180 degrees.

    [tex] E - E_{o} = -|\vec{F_{f}}|\vec{r}| [/tex]

    Ok now analysing forces on the y-axis

    [tex] N = mgcos \alpha [/tex]

    applying

    [tex] F_{f} = \mu N [/tex]

    [tex] F_{f} = \mu mgcos \alpha [/tex]

    [tex] E - E_{o} = -\mu mgcos \alpha |\vec{r}| [/tex]

    analyzing the first mechanical energy, we got only kinetic energy, because we put our initial distance point there, so our gravitational potential energey is 0

    [tex] E - \frac{1}{2}mv^2 = -\mu mgcos \alpha |\vec{r}| [/tex]

    Now for the final energy, we know the object speed is 0, therefore kinetic energy is 0, and it only has gravitational potential energy. The distance for the gravitation potential energy can be obtained with a little geometry (trigonometry).

    [tex] sin \alpha = \frac{y}{d} [/tex]

    [tex] dsin \alpha = y [/tex]

    where d is the hypotenuse and the distance travelled, and y is the distance for the Gravitational potential energy.

    Substituiting back in our equation.

    [tex] mgdsin \alpha - \frac{1}{2}mv^2 = -\mu mgcos \alpha |\vec{r}| [/tex]

    Solving for d

    [tex] d = \frac{\frac{1}{2}v^2}{(gsin \alpha + \mu gcos \alpha)} [/tex]

    Remember

    [tex] |\vec{r}| = d [/tex]

    Oh and btw [itex] d \approx 1.54 m [/itex]
     
    Last edited: Oct 9, 2004
  15. Oct 9, 2004 #14
    okay, thank you SOOOO MUCH. i was seriously going to go crazy.... one thing though, i'm getting 1.6... did you get exactly 1.5?
     
  16. Oct 9, 2004 #15
    nevermind, i'm stupid. THANKS AGAIN!
     
  17. Oct 9, 2004 #16

    Pyrrhus

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    I shouldn't have done that, but sometimes seeing how the problem is solved can be very good for enlightening (sp?). Just remember if you're stuck is there any geometry theorem to apply? or any a physics theorem?, and sometimes having a nap is good :smile: . By the way, you're welcome.
     
    Last edited: Oct 9, 2004
  18. Oct 9, 2004 #17

    Pyrrhus

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    By the way, try doing it the alternate way.

    Use Newton's 2nd Law

    [tex] \sum_{i}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    and some kinematics

    [tex] v^2 = v_{o}^2 + 2a(x-x_{o}) [/tex]
     
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