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PE, Work, object on inclined plane resting on a spring.

  1. Jul 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A block with mass m = 2.66 kg is placed against a spring on a frictionless incline with angle θ = 37.0° (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 14 N/cm, is compressed 19.1 cm and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_43.gif

    2. Relevant equations
    W=Fd, U(x)=1/2 kx2, U(y)=mg(Δy)

    3. The attempt at a solution
    Im on part b, im trying to solve U(y) but i dont have distance traveled in y direction. I think i need to find velocity next.
    20150702_191420_zpsvwnzsopp.jpg
     
  2. jcsd
  3. Jul 2, 2015 #2
    Are you incapable of doing something or do you want to see someone else doing it?
     
  4. Jul 2, 2015 #3
    Its more of a combination of both of those, but im looking for help to understand the technique in solving problems.
    This being my first semester of physics, ive noticed that solving physics problems is like working on a car, the first time you do something it takes 10 times longer than the second. All the problems ive posted on here are different concepts than what i have posted before.
     
  5. Jul 2, 2015 #4
    I have a feeling you know of an obvious solution to this problem, do you have any hints?
     
  6. Jul 2, 2015 #5

    haruspex

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    In part a) it says 19.1cm, but in your working you rounded it to 19cm.
    Your notes to the right of that show F=W/d. That only works for a constant force acting over distance d. If you want to know the compression force in the spring (but you don't need it for this question), use F=kx.
    For part b), think about energy, and how you can use the result from part a).
     
  7. Jul 2, 2015 #6
    for part c, im getting a distance of .612m, which is the distance from the release point along the inclined plane of which the object traveled. Using the formula U(x) =mg(x-x0) and solving for (x-x0). where, U(x)=25.53J, m=2.66kg, g=-15.69N
     
  8. Jul 2, 2015 #7

    haruspex

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    Why that value, and how come units of Newtons?
     
  9. Jul 2, 2015 #8
    because its the component of weight that alone the axis of the plane.
     
  10. Jul 2, 2015 #9

    haruspex

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    Then you mean ##F=mg\sin(\theta)=-15.69N##. But how does that lead to .612m?
     
  11. Jul 2, 2015 #10
    I thought it would make more sense to use it instead of -9.8m/s/s because the object is on an incline. I tried plugging in 9.8 but that still gives the wrong value. I tried using the kinematic displacement formula but time is not known. I have the objects initial velocity at release but im not sure how to get distance from that.
     
  12. Jul 2, 2015 #11
    i dont see why this wouldnt work;
    W=Fd
    d=W/F
    d=(-25.53)/(-15.69)
    d=400.6m​
     
  13. Jul 2, 2015 #12

    haruspex

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    By all means use ##g\sin(\theta)## instead of g, but don't write "g=" that value. g doesn't change. You compounded the error by incorporating the mass, turning it from an acceleration into a force. That's why I suggested you meant "F=". All very confusing.
    ,
    Your calculator appears to be broken.
     
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