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Finding Spring Constant/Compression

  1. Oct 1, 2014 #1
    We have online homework that checks our answers when input. I'm pretty confident about this problem, but it won't take my answers.

    1. The problem statement, all variables and given/known data

    A student proposes a design for an automobile crash barrier in which a 1350-kg sport utility vehicle moving at 25.0m/s crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00g.

    a) Find the required spring constant k. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground.

    b) Find the distance the spring will compress in slowing the vehicle to a stop.

    2. Relevant equations

    W = (1/2)kx^2 (work done by spring)
    v^2 = vo^2 + 2a(x) (to find x given a)

    3. The attempt at a solution

    I used 5.00(g) = 5(9.8) = 49 to find the maximum acceleration. I plugged this into v^2 = vi^2 + 2a(x) and got x = 6.38 m. This should be the answer to b), but the online homework program says it is wrong.

    I see no way to solve a) without b). I used W=(1/2)kx^2 (work done by a spring), set equal to (1/2)mv^2 (the kinetic energy of the car). Solving for k and plugging in values, I got 20731 N/m. The online homework program also rejected this answer.
     
  2. jcsd
  3. Oct 1, 2014 #2

    collinsmark

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    Hello BobBarker444, Welcome to Physics Forums! :)

    For what it's worth, I came up with the same answers that you did (ignoring any minor rounding differences).
     
  4. Oct 2, 2014 #3
    You cannot use ##v^2 = v_o^2 + 2ax##, as acceleration is not constant. (Recall derivation of SUVAT equations.)

    Equation for part (a) Looks alright.

    Equation for part (b): Hint - use ##F = ma = kx##

    Code:

    Code (Text):
    Solve[{1/2*1350 * 25^2 ==  1/2 k x^2 && 5*9.81 ==  k x/1350}, {k, x}]
    Using Mathematica, I obtained ##k = 5196.75 N m^{-1}##, ##x = 12.7421 m##.
     
    Last edited: Oct 2, 2014
  5. Oct 2, 2014 #4

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    Doh! o:) <slaps self on forehead>

    Yes, like unscientific said, the kinematics equations for uniform acceleration do not apply here. 'Looks like we both made that mistake, BobBarker444.

    However, you shouldn't need to use Mathematica to solve this. Just realize that you have two equations, [itex] ma = kx [/itex] and [itex] \frac{1}{2} kx^2 = \frac{1}{2}mv^2 [/itex]. And you have two unknowns, [itex] k [/itex] and [itex] x [/itex]. That's enough to solve the problem with algebra.

    Note that the value of the acceleration, [itex] a [/itex] varies when the car is still moving and the spring has not reached its maximum compression. We know that the force on the car, thus the car's acceleration, increases the more the spring is compressed. But when the spring reaches the compression necessary to cause the acceleration to be 5g, we do not need to treat [itex] a [/itex] as a variable: we ask ourselves, "what is the [itex] x [/itex] that causes [itex] a = 5g [/itex], when using the [itex] ma = kx [/itex] equation?" So [itex] a [/itex] is not an unknown. The only unknowns left are [itex] x [/itex] and [itex] k [/itex].
     
    Last edited: Oct 2, 2014
  6. Oct 19, 2014 #5
    Haha, I am generally lazy so I used mathematica to solve it.
     
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