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Determining Spring Constant

by csgirl504
Tags: constant, determining, spring
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csgirl504
#1
Oct26-11, 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

v2=v02 + 2a(x-x0

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/s2
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.
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daveb
#2
Oct26-11, 12:11 PM
P: 926
It may be a matter of units or significant figures if you're entering it into a computerized Q&A set.
Liquidxlax
#3
Oct26-11, 12:35 PM
P: 321
edit i got 10121

aftershock
#4
Oct26-11, 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
I like Serena
#5
Oct26-11, 04:39 PM
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Hi csgirl504!

I'm afraid your relevant equation does not apply
v2=v02 + 2a(x-x0)
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.
aftershock
#6
Oct26-11, 07:04 PM
P: 113
Quote Quote by I like Serena View Post
Hi csgirl504!

I'm afraid your relevant equation does not apply
v2=v02 + 2a(x-x0)
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.
Right, I read maximum acceleration as average acceleration.
csgirl504
#7
Oct27-11, 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!
gneill
#8
Oct27-11, 10:06 PM
Mentor
P: 11,614
Quote Quote by I like Serena View Post
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.
Lowest point? Presumably the car is traveling horizontally. Why the mgl (gravitational PE) term?

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.
I like Serena
#9
Oct28-11, 05:23 AM
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Quote Quote by gneill View Post
Lowest point? Presumably the car is traveling horizontally. Why the mgl (gravitational PE) term?
Of course. That's much easier. :)
I don't know why I thought the car was falling.
gneill
#10
Oct28-11, 05:59 AM
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P: 11,614
Quote Quote by I like Serena View Post
Of course. That's much easier. :)
I don't know why I thought the car was falling.
Perhaps you've been hypnotized by watching too many bobbing mass-springs!
I like Serena
#11
Oct28-11, 06:16 AM
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Quote Quote by gneill View Post
Perhaps you've been hypnotized by watching too many bobbing mass-springs!
Aaah, the surprising slinky ehild introduced!
I had quite some success with it outside PF.


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