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(g is assumed as 10 $ms^{-2}$)

We lift a stationary object of mass, $m$, to a height $s$, ensuring it has got a resultant upward acceleration of 2 $ms^{-2}$. The required force to be applied by us for the process is ($12m$). At $s$, in order to bring it to a stop, we apply a force of ($2m$) in downward direction. And thus it stops and remains at rest at height $s$.

By work energy theorem, net work done is zero as kinetic energy is zero.

Net work done = [Work done by gravity] + [Work done by us] = 0

[-($mgs$)] + [($12ms$) - ($2ms$)] =0

which is mathematically right.

The question is in the very last term.

The downward force of 2 $m/s$ does not act for a distance $s$. Why is then the displacement $s$ taken there ? Or is the equation wrong ? What, then, is the right equation?

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  • $\begingroup$ I don't know where to begin on this question. The work/kinetic-energy theorem generally applies to horizontal displacements. Work is equivalent to energy. When there is a vertical displacement, net work has been done in order to increase the gravitational potential energy of the object that was elevated. In your problem statement, conservative forces are involved, so the work that is done is path independent. My conclusion: your concepts are inappropriately mixed together and conservation of energy has been totally ignored. Carefully re-read your physics book. $\endgroup$ – David White Apr 21 at 18:30
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    $\begingroup$ @DavidWhite The OP is confused, but not that confused. Meanwhile your comment confuses me. Are you claiming the work-kinetic energy theory only holds when displacements are horizontal?! $\endgroup$ – knzhou Apr 21 at 19:47
  • $\begingroup$ No. W/KE theorem is normally thought of in a horizontal reference frame, but the problem is not always horizontal. $\endgroup$ – David White Apr 21 at 19:55
  • $\begingroup$ What is a "horizontal reference frame"? A Cartesian reference frame has three axes of coordinates. Which one is horizontal? $\endgroup$ – nasu Apr 22 at 2:50
  • $\begingroup$ @Zam If you start to apply the downward force when the object is at height s, it won't stop at s (and it won't remain at s). You need to reformulate your problem. As it is, it does not make sense. $\endgroup$ – nasu Apr 22 at 2:52
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First of all, the work done in bringing the mass to a rest at height $s$ is done by gravity, not us. So your equation should be rearranged as follows.

$$[-mgs-m2s]+m12s=0$$

But I think there is a bigger problem with the equation. Gravity has to be able to do enough negative work such that the negative change in kinetic energy gravity does equals the positive change in kinetic energy that you gave the body by applying a force of $m12$ for whatever distance you applied to it. The max possible negative work done by gravity is $-mgs$.

The only way that can happen is if you stop applying that force short of the distance $s$. The greater the distance you apply it, the greater the negative work done by gravity needs to be. If the distance is too close to $s$, the negative work of gravity may not be enough.

Bottom line is the positive net work done by you must equal the negative net work done by gravity, or

$$-mgs (gravity)+mgs (you)=0$$ in order for the body to come to rest.

Hope this helps.

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