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Spring compression distance from a block sliding down a ramp

  1. Oct 21, 2006 #1
    A 10kg box slides 4.0m down a frictionless ramp, then collides with a spring whose spring constant is 250 N/m. (Also, the ramp is angled at 30 degrees).
    a). What is the maximum compression of the spring?
    b). At what compression of the spring does the box have its maximum velocity?

    I started the problem by finding the final velocity of the block at the end of the 4m distance (which is the beginning of the spring) by finding the acceleration ( a=9.8*sin(30) ), then I applied it to the equation Vf ^2= Vi ^2 * 2a(delta x), and recieved an answer of Vf=6.261 m/s. I used this result for the initial velocity of the block moving down on the spring. After that, I used the equation

    .5*m*(Vf^2)+.5*k*(Xf - Xequilibrium)^2 +mgXf =.5m(Vi^2) + .5k(Xi - Xequilibrium)^2 + mgXi

    I considered Xi = Xequilibrium = 0; Vf = 0; and Vi =6.261
    but got the wrong answer (0.88m). The correct answer is 1.46m. I'm not really sure how to approach part b either.
  2. jcsd
  3. Oct 21, 2006 #2


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    Homework Helper

    Use energy conservation from the point at which the ball meets the spring to the point of maximal contraction of the spring, it gave the right answer.
  4. Oct 21, 2006 #3
    Thats what I tried with the equation, I guess my real problem lies with how I set it up... I went about it like this

    (final kinetic energy)+(final elastic energy)+(final gravitational potential energy) = (initial kinetic energy)+(initial elastic energy)+(initial gravitational potential energy)
    so I got the equation (each segment is the respective energy)
    .5*m*(Vf^2)+.5*k*(Xf - Xequilibrium)^2 +mgXf =.5*m*(Vi^2) + .5*k*(Xi - Xequilibrium)^2 + mgXi

    I considered the initial velocity of the block beginning to compress the spring as 6.261, The final velocity of the block as 0 (since it would be motionless at the maximum compression). I also considered the initial length of the spring without being compressed to be 0, as is equilibrium length. That way I could cancel out ".5*m*(Vf^2)", ".5*k*(Xi - Xequilibrium)^2", and "mgXi", which leaves
    .5*k*(Xf^2) + mgXf = .5*m*(Vi^2),
    giving me a quadratic equation, which I took the positive number for the position. I tried again, but no luck still... my question is, how do I set the equation up correctly?
  5. Oct 21, 2006 #4
    You did not account for your inclined plane.

    Its going to be pushing up the incline, and I'm pretty certian there is some trig work involved.
  6. Oct 21, 2006 #5

    Doc Al

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    Staff: Mentor

    Xf is measured along the incline, but potential energy (mgy) uses the change in height (vertical distance). (This is what GoldPheonix was talking about.)

    General comment: Nothing wrong with your overall approach (except for that error above), but you could have saved a bit of work by using conservation of energy right from the start. The initial energy is gravitational PE, the final energy spring PE (assuming you set your zero point at the lowest position). No need to calculate the velocity when it hits the spring.
  7. Oct 21, 2006 #6
    Ooooh I got it now, thanks much for all of your help folks :biggrin:
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