The work done against gravity is $mgh$, well at least that's what my textbook says. I have a question: I can apply a force say 50N, so total work done = $mgh + mah$. Where $ma$ = Force. But the truth is irrespective of the force applied, the work done against gravity is always $mgh$. Why?

For example, when I move an object with a force, the work done is more, so work depends on the Force. But in case of gravity it always depends upon the weight

  • $\begingroup$ Um...because $F_g = mg$ is the force of gravity? $\endgroup$
    – ACuriousMind
    Sep 12, 2014 at 14:11
  • $\begingroup$ But Why don't we add the force which I'm applying? $\endgroup$ Sep 12, 2014 at 14:13
  • $\begingroup$ Explain the downvote! Why? $\endgroup$ Sep 12, 2014 at 14:33
  • $\begingroup$ Related: physics.stackexchange.com/q/130690/2451 $\endgroup$
    – Qmechanic
    Sep 12, 2014 at 14:47
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    $\begingroup$ That is work done against a force, this is against gravity. Come up with a better reason and stop down voting without a valid reason. $\endgroup$ Sep 12, 2014 at 14:52

2 Answers 2


If I take a mass $m$ and apply a force $F$ (greater than $mg$) to it for a distance $h$ upwards then I will do work of:

$$ W = Fh \tag{1} $$

The force $F$ has to be greater than the force due to gravity, $mg$, or the object won't move upwards, so let's write the force I apply to the mass as:

$$F = mg + F'$$

then equation (1) becomes:

$$\begin{align} W &= (mg + F')h \\ &= mgh + F'h \end{align}$$

and the first term $mgh$ is work done against gravity while the second term is the work done to increase the velocity of the mass i.e. after the distance $h$ the velocity of the object will be given by:

$$ \tfrac{1}{2}mv^2 = F'h $$


$$ v = \sqrt{\frac{2F'h}{m}} $$

So if you apply any force $F$ over a distance $h$ then subtract off the increase in the kinetic energy you'll be left with an amount of energy equal to $mgh$. That's why the work done against gravity is always $mgh$.

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    $\begingroup$ what do you mean by work against gravity? works is done on objects not forces. $\endgroup$
    – user65081
    Oct 24, 2014 at 20:35

You might want to change your question title to "Work done by gravity," because that is what is implied by the variables mgh. Of course, you can add a greater force than that of gravity, which would cause whatever object to which the force is applied to accelerate (since the forces are not in balance). No matter how much force you apply, the force of gravity (ignoring general relativity or changes in g by location or elevation) does not change, thus the work done BY gravity does not care about how much force you exert. Does that help?

  • $\begingroup$ So is there no concept like work done against gravity $\endgroup$ Sep 12, 2014 at 14:41
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    $\begingroup$ @AkashLakshmanan, I guess I would not phrase it like a fight. Rather, I would phrase it in terms of what participated and how that "what" participated. So in your example, it would be more appropriate to say that "you" applied a force of 50 J to raise an object to a height, h. Thus, you did work on the object, not against another force. This becomes important when you talk about engines, where a system can do work or have work done on it. $\endgroup$ Sep 12, 2014 at 14:49
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    $\begingroup$ Yes, and see comment on Rennie's answer. Work is done on objects not forces. In this case gravity is just a force acting on the object, and it does negative work if the object goes up. $\endgroup$
    – user65081
    Oct 24, 2014 at 20:37

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