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Many introductory physics books just write that potential energy of a body of mass $m$ at a height $h$ as $U_\text{g}=mgh$. However, they never show how this was derived. I'm interested in knowing this derivation – if possible, avoiding calculus.

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    $\begingroup$ potential energy is defined using calculus. $\endgroup$ – AccidentalFourierTransform Mar 3 '16 at 17:40
  • $\begingroup$ Vaguely related: physics.stackexchange.com/q/122767/50583 I don't really get what you're asking here - if you don't know calculus, how can you define potential energy, which is just the integral of a conservative force? $\endgroup$ – ACuriousMind Mar 3 '16 at 17:48
  • $\begingroup$ Hint; if you do not know calculus then divide the height in n-parts and mark those points as y1,y2,...... calculate the work done in elemental paths and try adding it. $\endgroup$ – drvrm Mar 3 '16 at 19:11
  • $\begingroup$ @ Tweej and Tatan- Don't you think that work done by gravity will be -mgh rather than mgh? $\endgroup$ – Parth Mar 4 '16 at 0:59
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Suppose I let an object fall from rest for a distance $h$. Given that the gravitational acceleration is $g$ the velocity of the object will be given by the SUVAT equation:

$$ v^2 = u^2 + 2gh $$

In this case the initial velocity $u=0$ so we just get $v^2 = 2gh$. The kinetic energy of the object is given by:

$$ T = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m(2gh) = mgh $$

If energy is conserved the increase in kinetic energy must be equal to the decrease in potential energy, so we get:

$$ \Delta U = -\Delta T = -mgh $$

This tells us that if we lower the object by a distance $h$ the potential energy decreases by $mgh$, and conversely that if we raise it by a distance $h$ the potential energy increases by $mgh$.

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At the risk of suggesting an overly simple answer, work done on a body is defined as $W = F \cdot d$.

We know that the force of gravity acting on a body is $mg$.

The perpendicular distance it travels is $h$, so $$ W=E=F\cdot d= mgh\,. $$

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By definition we know that-

Gravitational Potential Energy of a body is the work done against gravity in raising it to a certain height h.

We have$$ \text{work}~~=~~\text{force}~\times~\text{displacement} \,.$$

Here force of the body is the weight acting vertically downwards=$mg$ and displacement is $h.$

So,$$ \text{work} ~=~mg\cdot h ~=~mgh \,.$$

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The form $U=mgh$ is simply an approximation to allow people to quickly calculate small changes in the gravitational potential energy of the system. The actual value of $U_g$ is usually not important in classical mechanics. $\Delta U_g$ is the important concept.

On a planet, small changes in height change $U_g$ in a space where the gravitational force is almost constant, $\vec{F}=m\vec{g}$. A change in potential energy (in classical mechanics) is calculated by the negative work done by a force while the position changes: $$\Delta U_{y'\to y'+h}=-\int_{y'}^{y'+h} mg(-\hat{j})\mathrm\ {d}y $$ The integral is easy: $$\Delta U_{y'\to y'+h}=\left. mgy\right| _y^{y+h}= mg(y'+h-y')=mgh$$

We see that the actual starting point is unimportant as long as the gravitational field can be considered constant. $h$ is positive if the particle is moved opposite the gravitational field (up), and negative if the particle moves down.

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protected by AccidentalFourierTransform Jun 16 '18 at 17:51

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