Determining Spring Constant

In summary: Aaah, the surprising slinky ehild introduced!In summary, the homework equation does not apply when a is not constant. The car's maximum acceleration occurs when the spring is maximally compressed.
  • #1
csgirl504
18
0

Homework Statement




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?


Homework Equations



F=ma

v2=v02 + 2a(x-x0

F=-kx

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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  • #2
It may be a matter of units or significant figures if you're entering it into a computerized Q&A set.
 
  • #3
edit i got 10121
 
  • #4
Don't round your values for F or x

k will come out over more than +5 of what you originally got
 
  • #5
Hi csgirl504! :smile:

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.
 
  • #6
I like Serena said:
Hi csgirl504! :smile:

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.
 
  • #7
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
I like Serena said:
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.
 
  • #9
gneill said:
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.
 
  • #10
I like Serena said:
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! :smile:
 
  • #11
gneill said:
Perhaps you've been hypnotized by watching too many bobbing mass-springs! :smile:

Aaah, the surprising slinky ehild introduced!
I had quite some success with it outside PF. :wink:
 

1. What is a spring constant?

A spring constant, also known as a force constant, is a measure of the stiffness of a spring. It is represented by the symbol k and is measured in units of force per unit length (such as N/m or lb/in).

2. How is the spring constant determined experimentally?

The spring constant can be determined experimentally by using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This can be done by measuring the force applied to the spring and the resulting displacement, and then plotting a graph of the data to find the slope, which is equal to the spring constant.

3. What factors can affect the spring constant?

The spring constant can be affected by several factors such as the material of the spring, the thickness and length of the spring, and the number of coils in the spring. Additionally, the temperature and the amount of stress or strain applied to the spring can also impact its stiffness and thus its spring constant.

4. How is the spring constant used in real-world applications?

The spring constant is used in various real-world applications, such as in the design of suspension systems in vehicles, as well as in the manufacturing of items such as mattresses, trampolines, and pogo sticks. It is also used in physics experiments to study the behavior of springs and other elastic materials.

5. Can the spring constant change over time?

Yes, the spring constant can change over time due to factors such as wear and tear, temperature changes, and material fatigue. In some cases, the spring constant may also change if the spring is stretched or compressed beyond its elastic limit, causing permanent changes to its structure and stiffness.

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