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Homework Help: Spring Forced Distance

  1. Dec 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring of length 0.80 m rests along a frictionless 30° incline (a). A 2.6 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m. (b)
    http://img64.imageshack.us/img64/5889/physics.jpg [Broken]

    The mass is pushed down, compressing the spring an additional 0.60 m, and then released. If the incline is 2.0 m long, determine how far beyond the rightmost edge of the incline the mass lands.

    The spring constant is 127.4

    2. Relevant equations
    Us=1/2kx²
    v²=v0²+2ax
    v=v0+at
    x=x0+v0t+1/2at²

    3. The attempt at a solution
    0.8m, in a 30° angle, on the spring will count as the 0 point.
    The potential energy of the spring on the block is 1/2kx²=Us so
    (1/2)(127.4)(0.1+0.6)²=31.213J

    Ki+Ui=Kf+Uf
    0+31.213=Kf+0

    K=1/2mv²
    31.213=1/2(2.6)v²
    4.9=v

    http://img691.imageshack.us/img691/5889/physics.jpg [Broken]
    Diagram of forces show gravity acts against it so mgsin30
    (2.6)(-9.8)sin30=-12.74N
    F=ma
    -12.74=2.6a
    -4.9=a

    v²=v0²+2a(x-x0)
    v²=4.9²+2(-4.9)(1.9)
    2.32=v

    So the block leaves the ramp at ~2.32m/s, 30° above the horizontal.

    I use the time it takes for the block to reach its peak height and time to hit the ground and multiply that with the velocity it moves rightwards.

    v=v0+at
    0=2.32sin30+-9.8t
    0.12=t1

    In this time the block moves up so its final height is
    x=x0+v0t+1/2at²
    x=0+(2.32sin30)0.12+(1/2)(-9.8)(0.12)²
    x=0.06875+1 (original height of right edge of ramp)
    x=1.06875

    Next is the time it takes to land,
    x=x0+v0t+1/2at²
    1.06875=0+0+(1/2)(9.8)t²
    0.467=t2

    t1+t2=0.585

    Finally the distance it goes away from the edge is velocity times time so

    x=vt
    x=(2.32cos30)(0.585)
    x=1.1771557879568892088679537911192

    Is this right or are there something wrong with the way I approached the problem?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 25, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's the elastic potential energy at the lowest point.

    Don't forget gravitational potential energy.

    Hint: Rather than treating the spring and gravity separately, use conservation of energy to solve for the speed of the mass as it leaves the incline.
     
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