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In this stack-post, an equation relating the external work done and the change in potential and kinetic energy of the system was derived by the user Biophysicist. Given: $$ W_{ext} = \Delta K + \Delta P$$ However, it is an equation often used in books and videos that :

$$ E= P + K$$

How was the above energy equation derived? Or is it a definition?

It seems to be related to being the previous equation by the fact that $ \Delta E = W_{ext}$

We can see this equation at 24:09 of this video

Qmechanic
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tryst with freedom
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3 Answers3

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$E=P+K$ is just a mathematical identity.

You just define a function $P(x)$ such that $-\frac{dP}{dx}=m\frac{d^2x}{dt^2}$, where $x$ is a function of $t$.

Now you multiply both sides by $dx$:

$$-{dP}=m\frac{d(\frac{dx}{dt})}{dt}dx$$

Now you just re-name $\frac{dx}{dt}$ as another function $v$, and integrate both sides to get:

$$-(P_2-P_1)=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$

Or

$$P_2+\frac{1}{2}mv_2^2=P_1+\frac{1}{2}mv_1^2$$

Since this sum $P+K$ remains unchanged as you transition from state $1$ to state $2$, this $P+K$ is an important quantity named $E$.

The above is a mathematical fact and would work for abstract mathematical functions having nothing to do with the real world.

What makes it interesting in the real world is that nothing is saying that a $P(x)$ would exist with the property $-\frac{dP}{dx}=m\frac{d^2x}{dt^2}$, where $x$ represents the position function of particles. Mathematics doesn't guarantee the existence of $P(x)$. The fact that a potential function $P(x)$ exists for all fundamental forces makes energy interesting.

tryst with freedom
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Ryder Rude
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  • There is a lot of work to be done on the formatting of this answer. I've done some basic ones – tryst with freedom Oct 13 '20 at 13:03
  • @Buraian i thought it looked fine. Any formatting tips? And is this helpful? Btw you're right that $E=P+K$ is a definition. There's no other definition of $E$. $P+K$ gets a name because it remains conserved as I showed in the answer. – Ryder Rude Oct 13 '20 at 13:32
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Mechanical energy is conserved only when there are no non conservative forces present, i.e., when $W_{ext}$ is zero in your equation. In that case, $$\Delta K = - \Delta U$$ $$K_f - K_i = U_i - U_f$$ $$K_i + U_i = K_f + U_f$$

Because the initial and final points are completely arbitrary, their sum must be the same for all points and so $K+P$ is constant at all points and you call that constant the mechanical energy $E$ of the system.

Brain Stroke Patient
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  • I meant this in a general context but you seemed to have to answered in context of mechanical systems.. or was my statements only holding in mechanical systems? – tryst with freedom Oct 13 '20 at 13:02
  • If you keep track of everything and if you take your system to be big enough, then $W_{ext}$ should generally be zero because all the fundamental forces are conservative as far as I know. So this is also true if you, for example don't consider friction to be a non conservative force but rather the sum of a lot of small conservative forces. – Brain Stroke Patient Oct 13 '20 at 13:08
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It seems to be related to being the previous equation by the fact that $ \Delta E = W_{ext}$

And you would be correct.

If the defined system is isolated, i.e, it does not exchange mass or energy with its surroundings, then its total energy $E$ is constant and equals the sum of its potential energy $P$ and kinetic energy $K$, or

$$E=P+K$$

Furthermore since $E$ is constant then $\Delta E=0$ and

$$\Delta P+\Delta K=0$$

If external work is done on the system, it is no longer isolated meaning $E$ can change. Therefore

$$W_{ext}=\Delta P+\Delta K=\Delta E$$

Hope this helps.

Bob D
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