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Spring/Energy Problem

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A new event has been proposed for the Winter Olympics. An athlete will sprint 100 m, starting from rest, then leap onto a 20 kg bobsled. The person and bobsled will then slide down a 50 m long ice covered ramp, sloped at 20°, and into a spring with a carefully calibrated spring constant of 2300 N/m. The athlete who compresses the spring the farthest wins the gold medal. Lisa, whose mass is 40 kg, has been training for this event. She can reach a maximum speed of 12 m/s in the 100 m dash.

    How far will Lisa compress the string?


    2. Relevant equations

    (0.5)*k*x_i^2 + (0.5)*m*v_i^2 + mgh = (0.5)*k*x_f^2 + (0.5)*m*v_f^2 + mgh

    3. The attempt at a solution

    0 + 0 + (40+20)(9.81)(50sin(20)) = 0.5*(2300)*(x_f)^2 + 0.5(60)(12)^2 + 0

    after calculating everything, I'm getting

    x_f = 2.23 meters

    while the answer in the back is 3.2 meters. There's likely something in my solution that I missed. Could anyone point it out for me?
    Last edited by a moderator: Apr 23, 2017 at 8:48 AM
  2. jcsd
  3. Nov 14, 2007 #2
    Your equation in (2) looks to be saying x_i = x_f. That doesn't seem right.

    Edit: Also, you didn't square x_i in your solution.
  4. Nov 14, 2007 #3
    Corrected! Apologies for doing this again.

    I assumed that the spring is relaxed, thus x_i = 0. Is this flawed?
    Last edited: Nov 14, 2007
  5. Nov 14, 2007 #4
    No, it isn't. I was looking at the wrong term.
  6. Nov 14, 2007 #5
    I'm having problems seeing how you can say initial kinetic energy is zero here. If you're using the combined mass for mgh, I think you have to use 12 m/s for v_i.
  7. Nov 14, 2007 #6
    Don't I still need to use that same velocity to find the compression distance in the RHS? If I use it initially, both KE's will cancel.

    Edit: I do see your point that there does exist a v_i in the problem, which makes KE_i exist. Escapes me how to find it though.
    Last edited: Nov 14, 2007
  8. Nov 14, 2007 #7
    When the spring is at max compression, velocity is zero. If it weren't, you would still have energy to be putting into the spring. Basically, everything goes into the spring.
  9. Nov 14, 2007 #8
    Gotcha. So KE_i + PE_i = PE_spring_f ?

    (.5)(20+40)(12)^2 + (20+40)(9.81)(50sin20) = .5(2300)(x_f)^2

    x_f = 3.54 meters. Does this sound about right? A little higher than the projected book value.
  10. Nov 14, 2007 #9
    Yeah, I get the same value. I don't know, everything looks right.
  11. Nov 11, 2009 #10
    I had the biggest issue with this problem until I realized that I was forgetting about momentum at the beginning of the problem!

    m x v_i = (m + M) v_f

    40kg x 12m/s = 60kg x v_f

    v_f = 8 m/s

    now you should get the correct answer!
  12. Mar 25, 2010 #11
    Basically, you want to find the energy right before she gets to the spring. This is a two part problem:

    1) find vf in the equation .5m(vi)^2 + mgh = .5m(vf)^2 + mgh
    ("h" is zero on the right side at the bottom of the hill, so actually solve this equation for vf:)
    .5m(vi)^2 + mgh = .5m(vf)^2

    Remember, "vi", although it appears to be 12 m/s, is actually 8 m/s in this problem because you need to take momentum into consideration (as discussed above). if lisa (40kg) runs at 12 m/s and jumps onto the sled of 20 kg, then take the momentum (40kg*12 m/s) and divide by the new total mass (60 kg) to get 8 m/s.

    Now, you should solve for "vf" in the previous equation and get 19.97 (or just 20) m/s

    2) Now we can solve the next equation: the kinetic energy (.5m(vf)^2) right before Lisa hits the spring is equal to the spring's elastic equation .5K(xf)^2

    Solve for "xf": .5m(vf)^2 = .5K(xf)^2 (and remember, "vf" is 20 m/s)

    after all this, you should get about 3.23 m, as the answer in the first post states :smile:
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