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Complicated Incline Question

  1. Nov 24, 2007 #1
    1. The problem statement, all variables and given/known data

    Given the incline plane, see picture, a box is let loose, attached to a spring in a relaxed position, the box moves down 2.75 meters. The variables: incline is at 32 degrees, the friction coefficient is .125 and the mass of the box is 17.4 kg. The length it slides is 2.75 meters.


    Find the: spring constant (k) when friction is present?

    spring constant (k) when friction is not present?

    the velocity of the box when it is half way down the incline (so 1.375 meters)?

    2. Relevant equations

    The teacher said that I could use conservation of energy to solve this and said that I could set it up in the following way

    Work(in)= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-Vo^2)+(friction)(distance)+(1/2)(k)(change in spring position^2)

    I think that the Work goes to 0 , the Initial Velocity goes to 0, and the initial spring position is 0 ---- so----

    0= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-0)+(friction)(distance)+(1/2)(k)(change in spring position^2)

    Is there a less cluttered way of solving this problem????

    3. The attempt at a solution

    The initial height should be (2.75)(sin32)? Right?
    The (distance) for the friction, and spring should be 2.75 because that is how far it goes?
    Does the final velocity also go to 0, I am not sure about that?
  2. jcsd
  3. Nov 25, 2007 #2
    Does anyone have an idea on how to go about solving this question?
  4. Nov 25, 2007 #3

    Doc Al

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    Staff: Mentor

    Perhaps it will be easier to apply conservation of energy if you think of it like this:
    (initial mechanical energy) + (work done by friction) = (final mechanical energy)

    Mechanical energy is the sum of kinetic energy, gravitational PE, and spring PE. The work done by friction (where it exists) will be negative.

    To find the work done by friction, first find the friction force.
  5. Nov 25, 2007 #4
    Can anyone else get the same answers as me ---

    k with friction -- 92.1179
    velocity half way -- 3.587 m/s
    k without friction -- 105.277
  6. Nov 25, 2007 #5
  7. Nov 25, 2007 #6

    Doc Al

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    Staff: Mentor

    Why don't you show exactly how you got your answers.
  8. Nov 25, 2007 #7
    I have gotten to

    mgh=fs+1/2kx^2 ---- so ---- (17.4)(9.81)(2.75cos32)=(.125(17.4)(9.81)cos32)(2.75)+(1/2)(k)(2.75)^2
    SO k= 92.117


    I replug it into
    Work(in)= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-Vo^2)+(friction)(distance)+(1/2)(k)(change in spring position^2)
    To get a velocity of 3.587 m/s

    Then to get without friction I just take friction out of the equation to get
    mgh=1/2kx^2 where k then equals 105.277
  9. Nov 25, 2007 #8

    Doc Al

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    Staff: Mentor

    The change in height should be 2.75*sin32, not cos32.
  10. Nov 25, 2007 #9
    thank you I don't know what i was thinking there
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