Elastic potential energy - different methods, different results

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SUMMARY

The discussion centers on the calculation of spring constant (k) using two different methods: the velocity equation and conservation of energy. The user initially calculated k as 50,000 N/m using the equation v_{f}^2 = v_{i}^{2} + 2as, assuming constant acceleration. However, the correct spring constant is determined to be 100,000 N/m through the conservation of energy approach, which correctly accounts for the relationship between elastic potential energy and the square of the elongation of the spring. The error was identified as the incorrect assumption of constant force in the first calculation.

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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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