- #1
Is F=Kx Always Necessary for Spring Forces?
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SUMMARY
The discussion clarifies that the equation F = kx, which describes the spring force, is not always necessary for applied forces. It emphasizes that F can take any value independent of kx, particularly when considering scenarios where the mass is either stationary or moving at constant velocity. The participants highlight the importance of Newton's 2nd law in analyzing forces acting on a mass and conclude that equilibrium occurs when |F| = |kx|, while acceleration results when |F| ≠ |kx|.
PREREQUISITES- Understanding of Newton's 2nd law of motion
- Familiarity with spring force and Hooke's Law (F = kx)
- Basic knowledge of equilibrium conditions in physics
- Concept of mass and acceleration in dynamics
- Explore the implications of Newton's 2nd law in various force scenarios
- Study the conditions for equilibrium in mechanical systems
- Investigate the behavior of springs under different force applications
- Learn about dynamic versus static equilibrium in physics
Students of physics, educators explaining mechanics, and anyone interested in understanding the relationship between forces and motion in spring systems.
- #2
jbriggs444
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Are you trying to invoke Newton's 2nd law or Newton's 3rd law? What, if anything, do either have to say about the situation?navneet9431 said:
- #3
navneet9431
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I am not trying to invoke any law
jbriggs444 said:
- #4
sophiecentaur
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Is the mass stationary?
- #5
navneet9431
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No
sophiecentaur said:
- #6
russ_watters
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The way you've drawn the picture, it looks like F is a separate force applied to the mass, not the spring force. You'll have to tell us what the drawing means - you drew it! (presumably).navneet9431 said:
- #7
sophiecentaur
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Make the spring as weak as you like. Would F be zero?navneet9431 said:
- #8
navneet9431
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Yes,F(on right) is a separate force applied to mass m other than the spring force acting on the left side.
russ_watters said:
- #9
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Please write down for us your Newton's 2d law force balance equation on the mass M.
- #10
navneet9431
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F-kx=ma
Chestermiller said:
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Perfect. Does that answer your question?navneet9431 said:
- #12
russ_watters
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In that case it can have literally any value you choose to give it.navneet9431 said:
- #13
navneet9431
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So it means that F=/=kx
Chestermiller said:
- #14
navneet9431
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But is it possible that F=kx in any condition?
- #15
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navneet9431 said:
If the mass is stationary, or if F is applied in such a way that the mass is moving at a constant velocity. Look at the equation you wrote. What is "a" under these two situations?
Zz.
- #16
navneet9431
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Thanks
ZapperZ said:
- #17
Kaguro
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navneet9431 said:
YOU will choose that.
If you choose |F|>|kx|, then m won't be at rest, but will have a constant acceleration rightwards.
If |F| = |kx|, then you have achieved an equilibrium, nothing moves. (In lab, this is how we calculate spring constant, where F=mg
)If |F| < |kx|, then you will have the spring pulling m leftwards.
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