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According to the solution for this

It says work done by all the net forces on the block equals the potential energy of the spring at equilibrium and to calvulate work done it only considers the frictional force above the plane and the work done by gravity not work done by tension. Why tension does no work here ? Here the elongation of spring must be $0.1 m$ too.

But why is there no tension here won't the $1 kg$ block push the string and at first the spring will yield and then slowly it will oppose and tension should come.

Also why do we have to use energy analysis here why the answer comes wrong when I do by force analysis ie at equilibrium I equate force $kx$ to net force on block ie $mg sin(37°)$ - $f$ where $f$ is force of friction?

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    $\begingroup$ Could you avoid so many typos? It's a pain to correct them. $\endgroup$ – Kunal Pawar Apr 21 '17 at 16:34
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    $\begingroup$ ... and format equations with LaTex/MathJax? $\endgroup$ – NickD Apr 21 '17 at 16:36
  • $\begingroup$ @KunalPawar why is it so important for you to correct them ? Anyway thanks. $\endgroup$ – Matt Apr 21 '17 at 16:40
  • $\begingroup$ It's imperative that the question be framed in such a way that is intelligible to other users. If it isn't so, you might not get any answers simply because others failed to comprehend what you intended​ to ask... $\endgroup$ – Kunal Pawar Apr 21 '17 at 16:47
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Even though the laws of motion allow us to deal with a whole range of problems, they fall short when it comes to variable forces. Remember that all the kinematic concepts or the dynamical problems you dealt with so far had constant acceleration.

The problem you've shown has a spring. And the spring force is not a constant force. It's a very good example of a variable force governed by the relationship:

$\vec{F}(x)=-k\vec{x}$

Where $\vec{x}$ is the displacement of the spring from it's intial position.

Since the displacements of the spring (a stretch or a compression) are in one dimension. The formula is often written as

$F(x)=-kx$.

Note that writing the force as $F(x)$ brings out the fact that it is not a constant. It is a function of the extension or compression $x$ of the spring.

I hope you can now understand why you're getting a wrong answer when you're blindly equating stuff. When you follow the incorrect approach, you assume that the force is constant.

Hence the concepts of work and energy often simplify problems.

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  • $\begingroup$ Okay what about the work done by tension ? $\endgroup$ – Matt Apr 21 '17 at 16:52
  • $\begingroup$ The spring force exerts a force on the block and the weight of the block exerts a force on the spring which keeps it stretched. What tension are you trying to get at? $\endgroup$ – Kunal Pawar Apr 21 '17 at 16:56
  • $\begingroup$ the spring exerts force on the block via the string ie tension $\endgroup$ – Matt Apr 21 '17 at 17:06
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    $\begingroup$ If you are using the change in the potential energy of the spring as part of your energy conservation, that already accounts for the work done by the spring via the tension of the string. You can't count that energy change twice. It's either the work done by tension or the change in spring potential energy, not both. $\endgroup$ – Bill N Apr 21 '17 at 18:10
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The system consists of the spring and the 1kg block. The solution is looking at work done on the system by external forces, which are gravity and friction. Tension in the spring is an internal force.

The net work done on the system by the external forces equals the increase in the internal energy in the system. In this case there is only elastic PE because the block "comes to rest", but in general there could also be internal KE in the system if the block is moving.

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  • $\begingroup$ Tension is an internal or external force is a matter of choice. If I take only block as my system then it is external if both block and spring then it is intenal. $\endgroup$ – Matt Apr 22 '17 at 13:38
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    $\begingroup$ Yes, that is correct. You can choose to make the spring force an external force. Like friction, the spring does -ve work on the block. If you use this method of forces you will get the same answer. It is not true that we cannot use force analysis - in some cases it is easier to use the energy method. $\endgroup$ – sammy gerbil Apr 22 '17 at 13:46

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