Is F=spring force generated?

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  • #2
jbriggs444
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Is it necessary that the applied force F is equal to the spring force 'kx'?
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?
 
  • #3
navneet9431
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I am not trying to invoke any law
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?
 
  • #4
sophiecentaur
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Is the mass stationary?
 
  • #6
russ_watters
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View attachment 232217
Is it necessary that the applied force F is equal to the spring force 'kx'?
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).
 
  • #7
sophiecentaur
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  • #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.
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).
 
  • #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
Please write down for us your Newton's 2d law force balance equation on the mass M.
 
  • #12
russ_watters
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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.
In that case it can have literally any value you choose to give it.
 
  • #14
navneet9431
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But is it possible that F=kx in any condition?
 
  • #15
ZapperZ
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But is it possible that F=kx in any condition?
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.
 
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  • #16
navneet9431
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Thanks
If the mass is stationary, or if F is applied in such a way that the mass is moving at a constant velocity.

Zz.
 
  • #17
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But is it possible that F=kx in any condition?
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:biggrin:)

If |F| < |kx|, then you will have the spring pulling m leftwards.
 
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