Does Mass of Cart Affect Spring Extension?

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The discussion revolves around whether the mass of a cart affects the extension of a spring in a physics experiment. The user observed that the spring stretched more with the cart attached, leading to confusion about the underlying physics principles. Key equations indicate that at equilibrium, the forces must balance, and the spring force is proportional to its extension. However, it was clarified that the mass of the hanging object remains constant, meaning adding the cart should not affect the spring's extension. Ultimately, the conclusion is that the cart's mass does not influence the spring's stretch, as the critical factor is the hanging mass.
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Homework Statement



I'm writing a paper on one of my lab experiments and I'm not sure my physics concept is right? The question asks if the mass of a cart affects the extension of the spring. A spring is attached to a cart which is attached to sting through a pulley and connected to a hanging object of mass W (a picture is attached). From my results, the spring stretched slightly more with the cart than without the cart attached to the spring.

Homework Equations



I know that at equilibrium all forces equal 0. Σ Fx = Σ Fy = 0
Fs= kx
F=mg

(with cart)
Σ Fx = Fs –Th
Σ Fy = Tv + Fn - Mgw -Mgc

(without cart)
Σ Fx = Σ Fx = Fs –Th
Σ Fy = Tv - Mgw

The Attempt at a Solution



If k is constant and Fs= kx and F=mg then mg=kx. Then if the cart was attached, mass will increase, therefore x (the spring extension) will increase, which explains why the spring stretched more when the cart was attached.

Is this somewhat right, because he said i have the wrong concept? Can anyone help me? Am I missing something?
 

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kay123 said:
with cart)
Σ Fx = Fs –Th
Σ Fy = Tv + Fn - Mgw -Mgc

(without cart)
Σ Fx = Σ Fx = Fs –Th
Σ Fy = Tv - Mgw
This is a bit confused since it's not clear what object you are analyzing. You don't just add up all the forces that appear in a problem, you must pick an object (the cart or the end of the spring) and analyze the forces acting on it.
If k is constant and Fs= kx and F=mg then mg=kx.
Then if the cart was attached, mass will increase, therefore x (the spring extension) will increase, which explains why the spring stretched more when the cart was attached.
Here "m" is the hanging mass, which doesn't change when the cart is attached. So adding the cart would not affect the stretch of the spring.
 
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