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So i was given this question whereby there was an object on a plane inclined at 30 degrees to the horizontal going up the plane being pulled by a string attached to it which is also attached to another object held in the air from a smooth fixed pulley(creating a system). when using work-energy relation what my book has done is :

work done by net force = change in TME

ie W.D by driving force - W.D against resistance = change in TME

ie (0- W.D against resistance) = change in TME

my question is why is the tension not considered as the driving force here?

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2 Answers 2

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If you consider both objects and the string as your system then the tension in the string is an internal force, so it does no work.

If you consider one object on its own, then the tension in the string is an external force and does work, which may be positive or negative, depending on the direction in which the object moves. If one object moves up the inclined plane then the tension in the string does positive work on that object - but there is an equal and opposite amount of negative work done on the other object as it falls.

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  • $\begingroup$ thanks! but would you mind elaborating : '' If one object moves up the inclined plane then the tension in the string does positive work on that object - but there is an equal and opposite amount of negative work done on the other object as it falls.'' $\endgroup$ Mar 13, 2021 at 18:00
  • $\begingroup$ @ilovecupcakes This is just applying work done = force applied x distance moved in the direction of the force. Which part is not clear to you ? $\endgroup$
    – gandalf61
    Mar 13, 2021 at 18:16
  • $\begingroup$ nevermind, however i'd like to ask you this : in a system, you must treat objects seperately to find the tension, so does that mean the tension does not exist when you consider the whole system? $\endgroup$ Mar 13, 2021 at 18:23
  • $\begingroup$ oh, it's an internal force. $\endgroup$ Mar 13, 2021 at 18:30
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Total mechanical energy actually depends on what you choose as your system, I am pre assuming that you are only considering the object that is on inclined plane and earth as your system because if we choose only object as system then we won't be able to define the notion of potential energy as the internal force i.e gravity won't count.

So applying Work energy theorem, we get:

$$W_{external}+ W_{internal}=\Delta K_{system}$$

Now negative of work done by conservative internal forces of the system is equal to change in potential energy $\Delta U$ of the system.

So, $$W_{external}=\Delta U_{system}+\Delta K_{system}$$

Now account for work done by all the external forces on this system which includes work done by tension as well as it isn't part of our earth-object system.

It maybe the case that your book choose to take something else as a system and then they would have applied work energy theorem.

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  • $\begingroup$ uhm, i'm studying maths and not physics, but this is kind of a physics question so i figured i would ask it here, could you please give a more math-appropriate answer ? tbh im not understanding any of this, and logically so. $\endgroup$ Mar 13, 2021 at 17:48
  • $\begingroup$ @ilovecupcakes Hello, these equations make sense if and only if we have properly defined what our system is! Your book might have chosen some other system that's why they aren't counting work done by tension, most probably they would have chosen both of those objects as a system that's why the net work done by tension on the system as a whole is being counted as zero. $\endgroup$ Mar 13, 2021 at 17:52
  • $\begingroup$ well, i considered the whole system. $\endgroup$ Mar 13, 2021 at 17:57
  • $\begingroup$ @ilovecupcakes if you considered the whole system then there is no external force on the system and change in mechanical energy would be zero. $\endgroup$ Mar 13, 2021 at 18:00

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