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phys: update energy work equation

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eggy 2020-11-03 10:56:29 -05:00
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@ -432,7 +432,7 @@ $$W=\vec{F}\cdot\vec{d}\cdot\cos\theta$$
When work is done **by** exactly one force, the above is formula is used. However, when calculating for work **on** an object, or **total work**, the sum of all forces ($\Sigma F$) should be used instead of $F$. When work is done **by** exactly one force, the above is formula is used. However, when calculating for work **on** an object, or **total work**, the sum of all forces ($\Sigma F$) should be used instead of $F$.
The **work-energy theorem** states that work is also equal to the difference in kinetic energy of an object. The **work-energy theorem** states that the **net work on** an object is also equal to the difference in kinetic energy of an object.
$$W=\Delta E_k = \frac{1}{2}mv^2$$ $$W=\Delta E_k = \frac{1}{2}mv^2$$
## 2.4 - Momentum and impulse ## 2.4 - Momentum and impulse