
#1
Oct2611, 10:07 AM

P: 19

1. The problem statement, all variables and given/known data
What should be the spring constant k of a spring designed to bring a 1260 kg car to rest from a speed of 88 km/h so that the occupants undergo a maximum acceleration of 5.0 g? 2. Relevant equations F=ma v^{2}=v_{0}^{2} + 2a(xx_{0} F=kx 3. The attempt at a solution First I converted 88km/h to 24.4 m/s Then I found the F, by using F=ma F= (1260) * (49) <changed acceleration from g to m/s^{2} So I got 61740 for F Then I decided to find x I used the velocity equation 0 = (24.4)^{2} + 2 * 49x Solving for x gave me 6.08 Then I plugged that into the F=kx equation and I got k = 10155 But this is wrong!! I even tried putting in 10155 What am I doing wrong? I don't understand. 



#2
Oct2611, 12:11 PM

P: 927

It may be a matter of units or significant figures if you're entering it into a computerized Q&A set.




#3
Oct2611, 12:35 PM

P: 321

edit i got 10121




#4
Oct2611, 12:35 PM

P: 113

Determining Spring Constant
Don't round your values for F or x
k will come out over more than +5 of what you originally got 



#5
Oct2611, 04:39 PM

HW Helper
P: 6,189

Hi csgirl504!
I'm afraid your relevant equation does not apply v^{2}=v_{0}^{2} + 2a(xx_{0}) It only applies when a is constant. I recommend using conservation of energy. If we assume the spring starts in a neutral position and expands to a length l, we have: Kinetic Energy before + Potential Energy before = Elastic Energy after + Potential Energy after or: [tex]{1 \over 2}m v_0^2 + m g l = {1 \over 2} k l^2 + 0[/tex] The maximum acceleration on the occupants is when the car is at the lowest point. (What is the formula for the maximum acceleration on the occupants?) From this you get 2 equations with 2 variables (l and k). Solving it yields the value of k. 



#6
Oct2611, 07:04 PM

P: 113





#7
Oct2711, 08:49 PM

P: 19

Thanks for all of your help! The homework is closed, so I can't see what the right answer is, but I will ask my teacher! And now I know how to work it if it shows up on the test! Thanks again!




#8
Oct2711, 10:06 PM

Mentor
P: 11,440

The maximum acceleration occurs when the spring is maximally compressed (so that the velocity is zero and the restorative force is greatest). Equate the original KE of the car to the PE of the compressed spring; all of the car's KE is transferred to the PE of the spring when it is brought to a halt. The PE of the spring depends upon k and x (the spring compression distance). The acceleration of the car depends upon the spring force, which in turn depends upon the spring compression distance. Solve for k in terms of a. 



#9
Oct2811, 05:23 AM

HW Helper
P: 6,189

I don't know why I thought the car was falling. 



#10
Oct2811, 05:59 AM

Mentor
P: 11,440




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