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Suppose I have to displace a body to a height $h$ and I applied force more than its weight. So, there will be an acceleration on the body. Then the body will reach the height $h$. Then I let the object to be there. I have done work more than $mgh$ because I have applied more force on it than its weight. Then my question is, why do we say it has potential energy $mgh$ only?

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  • $\begingroup$ Work is not solely dependent on the magnitude of force: $W=\int \mathbf{F}\cdot\ d\mathbf{r}$ $\endgroup$ – Quantumness Jun 24 '18 at 19:09
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If you apply a force with module greater than its weight, the body will accelerate and it will have a non-zero velocity. In particular

$$ E_{\text{body}} = \Delta E_{\text{body}} = \int \vec F \cdot \mathrm{d} \vec s = \int (-m\vec g + \vec F_{\text{extra}}) \cdot \mathrm{d}\vec s = U_{g} + \int \vec F_{\text{extra}} \cdot \mathrm{d} \vec s$$

In particular, you can write the energy of the body as the sum of 2 terms, the kinetic energy and the potential energy

$$ E_{\text{body}} = \frac{1}{2}mv^2 + U_g = U_g + \int \vec F_{\text{extra}} \cdot \mathrm{d} \vec s$$

Simplifying $U_g$ you get that the body has a non-zero kinetic energy, which simply means that your body will not be at rest if you push it with a force always greater than its weight, which is reasonable.

If I understood your question, the point is that the energy is not potential only but has other components.

To get a different interpretation, let's consider a similar case. Consider the same body at rest at a height $h$ over the ground. The gravitational potential energy is the kinetic energy that your body will have at the moment it hits the ground. If the body is already moving when it is at $z=h$, of course, it will have more energy when it hits the ground, but that's not potential energy, it's already kinetic.

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If when you say "then I let the object to be there" you mean the velocity of the object is zero when it reaches a height h, then the only energy it possesses is potential energy, mgh. It has no kinetic energy. Hope that helps.

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