What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of kmh so that the occupants undergo a maximum acceleration of 5.0 g?

Answer 1

Talking about abstract models: springs don't really do that, you would need a damper. Spring (ideal one) would gradually slow the car and then repel it in opposite direction with the exact same velocity and the acceleration/deceleration would be greatest when the car comes to temporary rest. Of course there's no ideal spring and real spring has also properties of damper, but then it cannot be easily described using single abstract value of k.

Also, when modeling crash event, you have to consider the length of deceleration. No damper or spring will save you, when there's no room for deceleration and you have to calculate it.

Let's consider an ideal spring model:

For deceleration to complete(in this case, car gets to momentary rest and then is repelled in opposite direction immediately), car has to transfer all of its kinetic energy(here the maximum)

Ek = mvstart2 / 2 to the spring.

Spring's maximum potential energy is: Ep = kx2 / 2, where x is maximum distance from spring's equilibrium.

We can see that mv2 = kx2, so

mv2 / x = kx.

We also have to take into account the maximum deceleration/acceleration:

mamax <= kx, which we can simplify to only the border case, where:

mamax = kx.

Comparing both equations:

mamax = mv2 / x, so

x = v2 / amax

So, we've established our minimum spring length.

We can put it into second equation, like:

mamax = kv2 / amax, so we get

k = mamax2 / v2, which is our k.

Since you neglected to provide velocity, the best number for k is therefore (1/v2)*2.9X106 kg*m2/s4