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Kinetic and Potential energy problem

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data
    A sled and its rider have a mass of 90 kg and a speed of 3 m/s at point A on the frictionless hill of Figure 5.25. At the bottom of the hill, the sled moves onto a rough surface with a coefficient of friction of .6. How far from the base of the hill will the sled move before stopping?


    2. Relevant equations
    KEf + PEf = KEi + PEi

    V^2= Vo^2 - 2ax

    KE= 1/2m(v^2)

    PE= mgy

    3. The attempt at a solution
    What I attempted to do was to find the speed at the end of the ramp by using the formula
    KEf + PEf = KEi + PEi
    so
    1/2(90)(9.8)((3)^2) + (90)(9.8)(3) = 1/2m(v^2)
    v= 7.67

    Then I used the velocity to find the total distance by using the formlula V^2= Vo^2 - 2ax
    so
    0 = 60.2176 - 2(54)(x)
    // I got 54 by multiplying the coefficient of friction by the normal force//
    when you solve for x you get .56 m. The real answer is 5.77 m
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 20, 2014 #2

    BiGyElLoWhAt

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    Just use conservation of energy. It's way cleaner and easier. Find the total energy at the bottom of the hill just before the rider enters the rough surface, and use E0 - W = Ef = 0 (because the rider lost all potential energy in coming down the hill which was converted to kinetic energy, and then the rider and sled have -W done on them by the surface)
     
  4. May 20, 2014 #3
    How do I find the distance the rider traveled then?
     
  5. May 20, 2014 #4

    BiGyElLoWhAt

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    Work is ##\int_{0}^{d} F \cdot dx = \int_{0}^{d} F dx cos(\theta)##
    Idk if you know how to do integrals or not, this is pretty simple. ##cos(\theta)## is a constant, and since friction acts opposite the velocity, theta is pi, and cosine theta is -1. F is a constant (##\mu N##) and ##\int_{0}^{d} dx = x|_{0}^{d} = d-0##

    put it all together and work done by friction is ##\mu N d (-1) = -\mu N d##

    Use that for W in conservation of energy, and solve for d.

    Hope that helps. BTW what level physics/math are you in?
     
  6. May 20, 2014 #5
    I do not understand integrals or calculus only algebra
     
  7. May 20, 2014 #6

    BiGyElLoWhAt

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    that's ok, you know conservation of energy though right?
    You know the kinetic energy at Point a, and you know the potential energy at point a. That means you know the total energy at point a. Since no work is done between a and just before the rider hits the rough patch, that means you know the total energy right before the rough patch, as energy is conserved.

    I gave you the work formula for friction, so take the total energy just before the rough patch and plug it into the conservation of energy equation. Then take the formula for work that I gave you (which should have come up in class, I'm sure), and plug that in. Then set the whole thing equal to the total energy at the end.
    *hint* What's the potential energy at the bottom of the hill, and what is the kinetic energy of the rider once he has stopped, which is what your looking for *un-hint*

    As to the algebra, I figured as much, that's why I worked it out for you. =]
     
  8. May 20, 2014 #7

    BiGyElLoWhAt

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    Ooo also, I noticed one slight flaw in your conservation of energy equation just now!

    The proper equation to use is this:

    ##E_{total_{i}} \pm \Delta W = E_{total_{f}}##

    Need that work in there bro.
     
  9. May 20, 2014 #8

    haruspex

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    Seems a bit low. Check your calculation.
    The mass is 90kg. What is the weight?
     
  10. May 20, 2014 #9
    OK I figured out this problem in a different way. My formula for the conservation of energy is correct I looked it up. The reason I got the question wrong is one I calculated the final velocity wrong its actually 8.23 m/s two I didn't use the right formula for the deceleration due to friction. The way for calculating the acceleration due to friction when using a kinematic equation like I did is the use the formula a= -μg which is the acceleration equals the coefficient of friction times the force of gravity. I used it and found the correct answer.
     
  11. May 21, 2014 #10

    BiGyElLoWhAt

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    That works, but I wouldn't rely too heavily on that energy equation. Think about it, is the total energy of the rider and sled the same as it was at point a? Definitely not. That equation only works if there's no work, but in this case there is, so it doesn't work for the friction part. Its good for from point a to the bottom of the hill, but after that its useless.
     
  12. May 21, 2014 #11

    haruspex

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    I think you're confusing energy conservation (always true) with work conservation. There was nothing wrong with the method in the OP, just the details.
     
  13. May 21, 2014 #12

    ehild

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    All good, but you need much less work if you solve the problem symbolically first.

    write mgh+1/2 mvi2=1/2 mvf2.

    You can multiply the equation by 2/m, and you get vf2=2gh+vi2=6*9.8+32=67.8, vf=8.23 m/s

    "54" is not acceleration but force. To get acceleration, you divide the force of friction by the mass. Its magnitude is μg. Vo2is the same as vf2=67.8.

    ehild
     
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