Conceptual Question on the Force of a relaxed spring

AI Thread Summary
When a moving cart collides with a relaxed spring, the force exerted by the spring is not constant but varies with compression. According to Hooke's Law, the force increases linearly with the amount of compression or extension of the spring. The equation F = -kx describes this relationship, where k is the spring constant and x is the displacement from the equilibrium position. The equation F = 1/2kx^2 refers to the potential energy stored in the spring, not the force itself. Understanding these principles is crucial for accurately analyzing the dynamics of the system.
Kibbel
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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


\DeltaP\rightarrow =\rightarrowFnet\Deltat

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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