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We have a spring of spring constant $k$. If we apply force $F$ on the spring,it elongates. The restoring of by hooked law $kx$ also develops on spring. Then a time will come when $kx=F$ i.e equilibrium position is obtained. But according to Newtons 3rd law, if we apply force $F$ on th3 spring,the spring will also pull by the same force $F$ on the contact point of us and the spring. And since our force $F$ is constant,the force by which the spring pulls us(the restoring force) is also constant. But again by hookes law,restoring force is a variable which changes with extension of spring. So where is the apparent fallacy we are making?

Qmechanic
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a_i_r
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    The forces in Newton's third law act on different objects. The force the spring pulls us with is not the restoring force. The restoring force acts on the end of the spring from the other parts of the spring. – Marius Ladegård Meyer Apr 13 '23 at 19:13
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    By Newton's 3rd law, the force applied to the spring increases linearly as the spring stretches at a constant rate (no acceleration), such that it matches the restoring force generated by the spring. This means that it's not possible to apply a constant force to the spring while stretching it. – David White Apr 13 '23 at 23:22
  • @DavidWhite Are you sure about that?What about a vertical combination of spring block system? A constant force which is the weight of the block acts on the spring. – madness Apr 14 '23 at 05:03
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    @madness DavidWhite is correct. The vertical spring block system is not a counter example. The gravitational force on the block is constant, but the force in the spring is not – Dale Apr 14 '23 at 11:10
  • I don't understand the question. "Then a time will come when $kx = F$ i.e. equilibrium position is obtained." $kx = F$ is always applies to an ideal spring. There is no "time will come". And I can't see what that has to do with equilibrium. – garyp Apr 14 '23 at 11:27
  • See also https://physics.stackexchange.com/questions/45653/given-newtons-third-law-why-are-things-capable-of-moving and the links within. – Chemomechanics Apr 15 '23 at 00:27

2 Answers2

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The thing that matters is that you think of the spring as a uniform and unified object but it's not. Every part of it does not behave as a single object and when it does( the equilibrium position) we can apply Newton's third rule easily and get the required result.

When we stretch it the part of the spring has to apply a force on the previous part and so on.that part is in equilibrium and when all parts are in equilibrium it's F=Kx.

T J
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  • Pardon me but what do you mean by equilibrium position? Is it the position $x_0$ where external force $F$ and spring force $kx$ are equal? – a_i_r Apr 15 '23 at 13:19
  • F=kx is the equilibrium position – T J Apr 16 '23 at 01:48
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So where is the apparent fallacy we are making?

The fallacy is thinking that you can apply an arbitrary force on an ideal spring at any length. Newton’s 3rd law means that the force applied on the spring is equal and opposite the force the spring applies. Hooke’s law means that the force the ideal spring applies is proportional to the extension. Taken together they imply that it is possible to apply a given force on an ideal spring only if the spring is at the corresponding extension.

Dale
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  • Pardon please but as you mentioned in the comment,David white said that it is not possible to apply a constant force on the spring. I do not understand why the vertical spring block is not a counterexample to you. The weight of the block is the external force applied on the spring. This external force is constant. But the the force applied by the spring on the block as per hooks law is not constant. David white said that the external force can't be constant whereas the gravitational force which is the external force is constant. – a_i_r Apr 15 '23 at 13:16
  • @a_i_r you said “The weight of the block is the external force applied on the spring” but that is not true. The weight of the block is the gravitational force acting on the block. It is applied to the block, not to the spring. The force on the spring is the tension, and it is not usually equal to the weight. – Dale Apr 16 '23 at 00:12