Elastic potential energy - different methods, different results

AI Thread Summary
The discussion centers on the calculation of elastic potential energy and the discrepancies between two methods. The initial approach used the velocity equation, leading to a spring constant of 50,000 N/m, which was deemed incorrect. The correct method applied conservation of energy principles, yielding a spring constant of 100,000 N/m. The error was attributed to the assumption of constant acceleration and force in the first calculation. Ultimately, it was clarified that elastic potential energy is proportional to the square of the spring's elongation.
Iamconfused123
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Homework Statement
A car of mass 1000 kg hits the spring with a speed of 10 m/s and condenses it by 1 m. If we disregard the friction, what is the resistance constant of the spring?
Relevant Equations
##E_p=\frac{kx^{2}}{2}##, ##E_k=\frac{mv^2}{2}##
Can someone please tell me where I am wrong?
I tried to solve the problem using velocity equation; ##v_{f}^2= v_{i}^{2} + 2as## and got a= 50m/s^2, F= 50 000N and therefore F=kx -> k=50 000N/m because dx=1.

But it's not correct. When I do it using conservation of energy I get 100 000N/m. Which is correct according to the solutions.
##\frac{mv^2}{2}=\frac{kx^2}{2}## -> k=100 000N/m.

Thank you.
 
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Iamconfused123 said:
Can someone please tell me where I am wrong?
Your first calculation is based on assumption of a constant acceleration / constant force. This assumption is incorrect.
 
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Hill said:
Your first calculation is based on assumption of a constant acceleration / constant force. This assumption is incorrect.
Thank you very much. Just figured, Ep is proportional to the square of the elongation of the spring.
 
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