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In summary, the conversation discusses the relationship between the applied force F and the spring force 'kx'. It is determined that F and 'kx' do not necessarily have to be equal, but can have any value depending on the situation. The equation F-kx=ma is mentioned as a possible force balance equation for the mass M. It is also mentioned that if F is greater than 'kx', the mass will have a constant acceleration, if they are equal, the system will be at equilibrium, and if F is less than 'kx', the spring will pull the mass in the opposite direction.

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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:Is it necessary that the applied force F is equal to the spring force 'kx'?

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jbriggs444 said: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?

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Is the mass stationary?

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No

sophiecentaur said:Is the mass stationary?

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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:View attachment 232217

Is it necessary that the applied force F is equal to the spring force 'kx'?

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Make the spring as weak as you like. Would F be zero?navneet9431 said:No

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russ_watters said: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).

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Please write down for us your Newton's 2d law force balance equation on the mass M.

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Chestermiller said:Please write down for us your Newton's 2d law force balance equation on the mass M.

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Perfect. Does that answer your question?navneet9431 said:F-kx=ma

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In that case it can have literally any value you choose to give it.navneet9431 said:Yes,F(on right) is a separate force applied to mass m other than the spring force acting on the left side.

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So it means that F=/=kx

Chestermiller said:Perfect. Does that answer your question?

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But is it possible that F=kx in any condition?

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navneet9431 said: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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ZapperZ said:If the mass is stationary, or if F is applied in such a way that the mass is moving at a constant velocity.

Zz.

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navneet9431 said: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)

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

Yes, the spring force is generated by a physical spring. This force is a result of the deformation of the spring due to an applied external force.

The spring force can be calculated using Hooke's Law, which states that the force is directly proportional to the displacement of the spring from its equilibrium position. The equation is F = -kx, where F is the spring force, k is the spring constant, and x is the displacement.

Yes, the spring force can be negative if the displacement of the spring is in the opposite direction of the applied force. This indicates that the spring is being compressed, rather than stretched.

No, the spring force does not change with the mass of the object attached to the spring. It is solely dependent on the displacement of the spring and the spring constant, which is a property of the spring itself.

Yes, the spring force can be greater than the applied force if the displacement of the spring is large enough. However, this is only possible up to a certain point, as the spring will eventually reach its elastic limit and become permanently deformed.

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