Does Mass of Cart Affect Spring Extension?

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SUMMARY

The discussion centers on the relationship between the mass of a cart and the extension of a spring in a physics experiment. The participant concludes that the spring stretches more when the cart is attached, based on the equation Fs = kx, where k is the spring constant and x is the extension. However, the analysis reveals a misunderstanding; the mass of the hanging object (m) remains constant regardless of the cart's presence, indicating that the cart's mass does not influence the spring's extension. The correct interpretation is that the spring's extension is solely dependent on the weight of the hanging mass.

PREREQUISITES
  • Understanding of Hooke's Law (Fs = kx)
  • Knowledge of equilibrium conditions in physics (Σ Fx = Σ Fy = 0)
  • Familiarity with forces acting on objects (e.g., tension, gravitational force)
  • Basic concepts of mass and weight in physics
NEXT STEPS
  • Review Hooke's Law and its applications in spring mechanics
  • Study the principles of static equilibrium in physics
  • Explore the effects of mass on force and acceleration (Newton's Second Law)
  • Investigate the role of tension in pulley systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators seeking to clarify concepts related to mass and force interactions in experiments.

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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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