Energy

Mechanical Energy

Conservation of mechanical energy
$m\overrightarrow{v}=f\left(x\right)$
By applying numerous transformations on this equation, you can derive:
$\frac{1}{2}m{v}^2=\int f\left(x\right)dx+C$

Where L.S = KE and R.S = work + the integration constant

Conservative Forces

If the work done by f(x) is independent from the path and only depends on endpoints, the force is a conservative force and can be derived from a function P(x)

$f\left(x\right)=\frac{-dP\left(x\right)}{dx}=P\left(x\right)=-\int f\left(x\right)dx$
Relating to the equation above…
$\frac{1}{2}m{v}^2+P\left(x\right)=C$
Where the first term is K.E and the second is P.E.

Potential Energy

Potential energy is the “potential” of an object, as in, the potential to do work. This is kind of like stored energy

The sum of P.E and K.E is always constant!
That being said $\Delta KE+\Delta PE=0$
Meaning energy is conserved.

Energy of Rotational Motion

Linked from Rotational Mechanics
Work done by moment M to rotate a body by theta.
$w={\int }_0^{\theta }Md\theta$
$M=I{\omega }^{\prime }$
${\int }_0^{\omega }I\omega d\omega ={\int }_0^{\omega }Md\theta$
Where L.S is KE, and R.S is work.
Now we can define KE as
$KE={\int }_0^{\omega }I\omega d\omega =\frac{1}{2}I{\omega }^2$
For rotational (pulley) questions, remember to include rotational KE in the energy equations.