- #1
MightyDogg
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Homework Statement
- 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?
Homework Equations
F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2
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?