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Max value of spring constant

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data
    You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1500 N will move with speed 2.2 m/s at the top of a ramp that slopes downward at an angle 23.0 degrees. The ramp will exert a 586 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 7.6 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring k_max that can be used in order to meet the design criteria.

    2. Relevant equations

    F = -kx
    PEs = 1/2 kx^2
    PEg = mgh
    KE = 1/2 mv^2

    3. The attempt at a solution

    1/2mv^2 + mgh = Ff*d + 1/2 kx^2
    kx = 586 N

    k = 586/x

    1/2(1500/9.8)(2.2)^2 + (1500/9.8)(7.6sin(23)) = (586*7.6) + 1/2 kx^2

    371.143 = 1/2(586/x)(x^2)
    x = 1.27 m
    k(1.27) = 586

    k = 461.4 N/m -- this apparently is not the right answer; what am i doing wrong?
  2. jcsd
  3. Mar 16, 2010 #2
    Draw a free body diagram of the final state when the crate is at rest against the spring. How many forces are there, and what are they from? Also, the sum of the forces will be 0 since everything is at rest.
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