1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook