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Energy with friction

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1.20 kg object slides to the left on a surface having a coefficient of friction of 0.250. The object has a speed of v_i = 3.00 m/s when it makes contact with a light spring that has a force constant of 60.0 N/m. The object comes to rest (briefly) after the spring has been compressed a distance d. The object is then forced towards the right by the spring and continues to move in that direction beyond the spring's uncompressed position, finally coming to a rest a distance D to the right of the unstretched spring. Find:
    A) The distance of compression d.
    B) The speed v at which the object is moving when it reaches the uncompressed position of the spring after compressing the spring.
    C) The distance D at which the object comes to a rest.

    2. Relevant equations
    ΔKE + ΔUg + ΔUe + fkd = Fexternald

    3. The attempt at a solution
    I split this into two parts. Part I that includes the compression of the spring until the object is at rest, and Part II that includes the part where the object is launched from the compressed spring.

    In part one I said that the Kinetic Energy initially is 1/2 mv^2 and the Kinetic Energy final is 0 because the object comes to a rest. The Elastic Potential energy initially is 0 and the Elastic Potential energy is 1/2 k*Δx^2

    My energy equation was 1/2mv^2 - μkmgΔx=1/2kΔx^2

    I dont know if setting this equation up is correct and I'm not sure if the distance I should use for the frictional force is Δx or some other arbitrary number.
     
  2. jcsd
  3. Apr 28, 2015 #2

    jbriggs444

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    The setup for part I looks entirely correct.

    Yes, as the spring is being compressed, energy is lost to friction and the amount lost will be given by the frictional force times Δx.
     
  4. Apr 28, 2015 #3
    How should I go about solving for Δx? I get a weird polynomial where I get multiple answers for Δx.
    Should I set up some kind of work-Force relationship?
     
  5. Apr 28, 2015 #4

    jbriggs444

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    The equation is quadratic in Δx. If you replace "Δx" with the more familiar symbol "x", can you arrange the equation into the standard form for a quadratic equation and apply the quadratic formula?

    Edit (addition):

    It can often be the case that a problem will result in a quadratic formula that offers two answers. In the usual case, one of those answers will be obviously non-physical. For instance, the result may be negative

    Without having worked out the details, I would confidently expect the spurious result for delta x in this case to be negative.
     
  6. Apr 28, 2015 #5
    I'm currently doing that, the only problem I'm now having is that this isn't the first time I've had to apply the quadratic formula during homework or an exam and never know whether or not I'm supposed to use -b - √b2-4ac or -b + √b2-4ac
     
  7. Apr 28, 2015 #6

    jbriggs444

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    You are doing excellent work and do not seem to need any help at all.

    The fact that the quadratic formula produces two answers used to bother me as well. What helped was to see that there is usually a different way of looking at the problem so that both answers make sense. For instance, in a projectile problem where there are two solutions for when the projectile lands, one of those times will generally be prior to launch. So if you view the projectile as having a trajectory that extends infinitely into both future and past, both solutions are viable.
     
  8. Apr 28, 2015 #7
    Ah I see now. For this instance I wanted a distance, and since distances are generally positive unless its displacement I added the square root. I got the correct answer!
    Thank you very much for your help.
    I have an exam in a few hours and you've helped put me at ease.
     
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