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Homework Help: A block on a frictionless slope slides down and compresses a spring

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Hello, folks. I know this same problem has been answered before here but the previous responses are just not doing it for me, sorry!

    An inclined plane of angle θ = 20.0° has a spring of force constant k = 495 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 = 2.37 kg is placed on the plane at a distance d = 0.318 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?

    Here is an image of such a setup.

    2. The attempt at a solution

    So basically I want to find the displacement in the spring given these values. What I do first is get the velocity at which the box hits the spring so I can get it's kinetic energy at that point.

    It starts at .750m/s but it's acceleration due to gravity is 9.8*sin(20 degrees) since it's on an incline and it has to travel .318m to hit the spring so I've got:


    .318=0+0.75*t+1/2*(9.8*sin(20 degrees))*t^2

    t=0.26951 seconds

    now that we have time let's get final velocity.


    v_final=0.75+9.8*sin(20 degrees)*0.26951

    v_final=1.653 m/s

    Now let's get the kinetic energy of the block when it hits the spring.



    Now let's get the potential energy on the spring



    495/2*x^2 J

    now let's solve


    x=0.114m. This is incorrect.

    Where have I gone wrong?

  2. jcsd
  3. Oct 25, 2011 #2
    Could you perhaps try and use hookes law to see how much it compresses directly? F=-kx, x=(2.37*9.8*sin(20))/(495)=.016? its been a while since I did physiscs so im not 100% sure..

    is the real answer really close to what you have? it could be a rounding error..
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