# Why does Hooke's law not work here?

1. Oct 20, 2015

### MightyDogg

1. The problem statement, all variables and given/known data
1. A 1200-kg car moving on a horizontal surface has speed v = 85 kmh when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
2. Relevant equations
F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2

3. The attempt at a solution
I tried to find the average acceleration to slow the car from 85kmh to 0. I used the formula vfinal^2=vinitial^2 + 2ax, where velocity initial is 23.6m/s and x is 2.2m. This gave me an acceleration of -126m/s^2. Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

However, I am supposed to use the conservation of energy principle where KE=PE. This gives the correct answer. Why does my method not work?

2. Oct 20, 2015

### Ray Vickson

I does not work because the acceleration is not constant. Hook's Law works, but not your expression $v_f^2 = v_i^2 + 2ax$.

Last edited: Oct 20, 2015
3. Oct 20, 2015

### Mister T

By Hooke's law I assume you mean $F=kx$ where $F$ is the magnitude of the force and $x$ is the distance?

In that formula $F$ is not the average force. It's the magnitude of the force when the spring is stretched (or compressed) a distance $x$.

If the force were constant, that would work, but the force is not constant. You could integrate the force, or use energy concepts.

4. Oct 20, 2015

### MightyDogg

Oh, okay it makes sense now. Thank you both very much.