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Spring constant and maximum acceleration

  1. Feb 15, 2007 #1
    What should be the spring constant k of a spring designed to bring a 1150 kg car to rest from a speed of 99 km/h so that the occupants undergo a maximum acceleration of 5.0 g?

    I am unsure of the formula to find the spring constant by determining a maximum acceleration of 5 g
  2. jcsd
  3. Feb 15, 2007 #2
    What have you tried so far? Can you think of any formulas that might be relevant?
  4. Feb 15, 2007 #3
    PE= 1/2 k x^2
  5. Feb 18, 2007 #4
    This is how you should do it:

    1- change the initial velocity to m/s
    2- calculate acceleration: 5*9.8m/s^2 = -49m/s^2: note acceleration is negative to stop the car.
    3- calculate the distance (x) over which the car come to halt: V^2= V^2 (initial)+2ax (V= final velocity = 0, x= distance, a=acceleration)
    4- now you have 'x', calculate 'k' like this: F=-kx and F=ma (where m=mass)

    Solution: m=1150kg, V(initial)=99km/h*(1000m/km)*(h/3600s)= 27.5m/s, V(final)=0, now:

    0 = (27.5)^2 + 2*(-49)* x, x= 7.7m

    now we can calculate 'k': ma = -kx (this come from F = -kx hook's law), so

    1150kg*-49m/s^2 = -k * 7.7m, k = 7318.1 N/m

    Note: I did this calculation quickly and I might have errors but the method should be all right.

    Good luck!
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