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If a force is provided to lift a system (let's say a box) vertically upwards and there is no resistive force, from my understanding (though I may be misguided),it is said that the work done on the box is equal to the gravitational potential energy that it has gained.

I also understand that the work done on a system is equal to the change in its mechanical energy which is a combination of its kinetic and potential (in this case gravitational potential) energy.

So my question is how can the work done on the box be equal only to the the gain in gravitational potential energy if, by definition, the work done on a system is equal to the change in mechanical energy?

Wouldn't the kinetic energy of the box increase as well?

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You are getting mixed up between work done by a single force and the net work done.

The net (total) work done on an object is equal to it's change in kinetic energy $$W_\text{net}=\Delta K$$

However, it is most likely that in your education of the box lifting example they are talking about the work just done by the lifting force as well as applying the additional assumption that $\Delta K=0$. Then we have $$W_\text{lift}+W_\text{grav}=0$$ Or in other words $$W_\text{lift}=-W_\text{grav}=\Delta U_\text{grav}$$

Therefore, the work done on the box by the lifting force is equal to the change in potential energy only if there is no change in kinetic energy.

In the box lifting example if $\Delta K\neq0$ then you need more information to determine $W_\text{lift}$. You either need to know the change in kinetic energy or the force $\mathbf F$ you apply so that you can find $W_\text{lift}=\int\mathbf F\cdot\text d\mathbf x$.

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  • $\begingroup$ Would you say that I most likely am referring to situations where the box is lifted with a constant speed (minimum lifting force) therefore no change in kinetic energy? $\endgroup$ – user3602727 Nov 18 at 1:20
  • $\begingroup$ @user3602727 Probably. That is usually how that example goes. Although it doesn't have to be. $\endgroup$ – Aaron Stevens Nov 18 at 2:02
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If a force is provided to lift a system (let's say a box) vertically upwards and there is no resistive force, from my understanding (though I may be misguided),it is said that the work done on the box is equal to the gravitational potential energy that it has gained.

That is correct provided the velocity of the box is zero at a given height so that it has no kinetic energy.That means at some point you must reduce the upward force to decelerate the box so that its velocity is zero when it reaches the desired height.

So my question is how can the work done on the box be equal only to the the gain in GPE if by definition work done on a system is equal to the change in mechanical energy?

It can be, provided that the velocity of the box is zero when it reaches the desired height. The change in mechanical energy must include any change in kinetic energy as well as any change in potential energy. The work done by the external force will only be equal to the gain in gravitational potential energy if the box has zero velocity when it reaches the desired height. If it does have a velocity $v$, then the box has kinetic energy of $\frac{mv^2}{2}$ in addition to gravitational potential energy of $mgh$ when it attains a height of $h$.

thank you it does help, but can you can confirm that the work done ON a system is equal to its change in mechanical energy or is it change in just its kinetic energy

Yes the work done on an object equals its change in mechanical (kinetic plus potential) energy, but the net work done on an object only equals the change in the kinetic energy component of mechanical energy. That's because there are two forces doing work when an object is raised, the upward force exerted by you in the same direction as the displacement of the object, and the force of gravity acting downward opposite the direction of the displacement. By definition work is positive when the force is in the same direction as displacement. Thus you are doing positive work, and gravity is doing negative work. The sum of the two is the net work done on the object.

If you are doing more positive work on the object than the negative work done by gravity, positive net work is done on the object and the change in its mechanical energy will equal the positive change in kinetic energy plus the change in gravitational potential energy. The positive change in kinetic energy is due only to the net work done on the object by you. The change in gravitational potential energy is due only to the work done by gravity.

If the object begins at rest and ends at rest there is no change in kinetic energy and the net work done on the object is zero, that is, you are doing the same amount of positive work as the negative work of gravity. In effect, gravity takes the energy of the work done by you on the object and stores it as gravitational potential energy of the earth-object system. The change in mechanical energy is only gravitational potential energy.

Hope this helps.

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  • $\begingroup$ thank you it does help, but can you can confirm that the work done ON a system is equal to its change in mechanical energy or is it change in just its kinetic energy $\endgroup$ – user3602727 Nov 18 at 1:30
  • $\begingroup$ @user3602727 See revision to my answer. $\endgroup$ – Bob D Nov 18 at 14:01

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