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Assume that we applied a force on an object so it moves at a constant speed upward, and we defined our system as the object only.

I was taught in my physics class that if we want to calculate the total work done on the system I will use the formula $W_{net}$= $\Delta K$ which is known by the Work-energy theorem. Moreover, it can be written as $W_{net}$= $W_{\text{conservative}}$ (i.e. work by gravity) + $W_{\text{non-conservative}}$ = $\Delta K$, so that $W_{net}$=0 because $ \Delta K$ is zero.

If we now considered the system to be the object along with the earth what will the evaluation of $W_{net}$ and $W_{ext}$ be? Please include equations in the answer. Also, in this case, is $W_{\text{conservative}}$ still be included in $W_{net}$?

Young Kindaichi
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Marwa
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  • By putting gravity in parenthesis where you did, are you saying the work of gravity if non conservative? – Bob D Mar 03 '21 at 15:41
  • This may help Clarification about what makes a system isolated – mmesser314 Mar 03 '21 at 15:56
  • I don't think your equation is correct. $W_g=-\Delta P.E$ , which is only defined for a system , in this case the object and the earth. – Bhavay Mar 03 '21 at 16:43
  • What I think is that in the first equation $W_g$ is also considered as an external force on the system. Also, we have one object in the system so we can't say that $W_g$ = $-\Delta PE$ because PE is associated with a system of at least 2 objects. However, in the second equation, $W_g$ is internal to the system, so I can use the definition you've mentioned. @Bhavay – Marwa Mar 03 '21 at 17:11
  • I put gravity in the wrong place. I've edited it. @BobD – Marwa Mar 03 '21 at 17:12

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