cgward said:
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
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!