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Spring Compression at kinetic energy = 0

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    An inclined plane of angle θ = 20.0° has a spring of force constant k = 460 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.51 kg is placed on the plane at a distance d = 0.324 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

    2. Relevant equations
    K = 1/2 mv^2
    Elastic Potential energy of spring = 1/2 kx^2

    3. The attempt at a solution
    So first I found the acceleration of the object:
    gsin(θ) = a

    Then I found the velocity of the mass at the time it hits the spring
    d = vt +1/2 at^2
    t is about .269593
    so velocity when the object hits the spring = v+at which is about 1.654

    Then I had: 1/2 mv^2 = 1/2 kx^2
    solving for x, I got about 0.122 meters
    However, the website to send the answer says I'm wrong. I have looked this up online, and I have substituted my values into other peoples equations and I still get 0.122
    Thanks in advance!
  2. jcsd
  3. Sep 23, 2014 #2


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    Think of the change of gravitational potential energy while the spring is being compressed.

    Last edited: Sep 23, 2014
  4. Sep 23, 2014 #3


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    Consider the involved energies - kinetic and potential energy of the block and the energy stored in the compressed spring.
  5. Sep 23, 2014 #4
    Okay so gravitational potential energy is mgy. Would y just be the distance of the object to the spring, or would it be the distance plus x (the distance the spring compressed). So either:
    mg(d+x) + 1/2 mv^2 = 1/2 kx^2
    mg(d) + 1/2 mv^2 = 1/2 kx^2
  6. Sep 23, 2014 #5


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    Neither. It is the vertical distance the mass travels from initial to final position (maximum compression in this case).
  7. Sep 23, 2014 #6
    Um, but isn't d+x the initial position to the position of maximum compression?
  8. Sep 23, 2014 #7
    Okay I got it!
    mg*sin(theta) * (d+x) + 1/2 mv^2 = 1/2 kx^2 works.
    Thank you guys for all of your help!
  9. Sep 25, 2014 #8


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    That's correct, yes.
    The gravitational potential energy and the kinetic energy at the top
    is converted to the stored elastic enregy of the spring when the block
    comes momentarily to rest at the bottom.
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