Conceptual Question on the Force of a relaxed spring

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SUMMARY

The discussion centers on the behavior of a relaxed spring when a moving cart collides with it. According to Hooke's Law, the force exerted by an ideal spring is directly proportional to its displacement from the equilibrium position, expressed as F = -kx. The potential energy of the spring, represented by the equation U = 1/2 kx^2, is distinct from the force equation. Therefore, the force on the cart is not constant but varies with the degree of compression or extension of the spring.

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Homework Statement


When running a moving cart into a relaxed spring, will the force of the spring be constant on the cart?

Homework Equations


[tex]\Delta[/tex]P[tex]\rightarrow[/tex] =[tex]\rightarrow[/tex]Fnet[tex]\Delta[/tex]t

or F = 1/2kx^2?

The Attempt at a Solution



my guess is that the more you compress or strech a spring, the stronger the opposing force
 
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Your conceptual explanation is correct. That is essentially Hooke's Law. The force caused by an ideal spring increases linearly with it's compression (or expansion).

However, F does not equal 1/2kx^2. That is the potential energy. F=-kx
 

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