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$F_{c}={\frac {mv^{2}}{r}}$

$A_{c}={\frac {v^{2}}{r}}$

$G_{F}={\frac {G\times M_{1}\times M_{2}}{r^{2}}}$

$G_{a}={\frac {G\times M_{1}}{r^{2}}}$

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$Work=\Delta KE+\Delta U_{g}+\Delta W_{f}$

$F\times \Delta x\times \cos \theta ={\frac {1}{2}}mv^{2}+mg\Delta h+\mu F_{n}\cos \theta \Delta x$

$\Delta KE={\frac {1}{2}}mv^{2}$

$\Delta U_{g}=mg\Delta h$

$W_{f}=\mu mg\sin \theta$

$U_{0}+KE_{0}=U_{1}+KE_{1}+W_{f}$

$U_{G}=-{\frac {GM_{e}m}{r}}$

$P={\frac {W}{t}}$

$Work=Joules=Nm={\frac {kgm^{2}}{s^{2}}}$

$Power=Watts={\frac {Nm}{s}}={\frac {kgm^{2}}{s^{3}}}$

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$\sin ^{2}x+\cos ^{2}x=1$

$f(x)=f(a)\times f(b)\Longrightarrow f'(x)=[f'(a)\times f(b)]+[f'(a)\times f(b)]$

$f'(x)=[f'(a)\times f(b)]+[f'(b)\times f(a)]$

$1hp=745.70Watts$

$a_{1}=a_{0}-{\frac {f(a_{0})}{f'(a_{0})}}$

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Work = Change in Kinetic Energy + Change in Potential Energy + Work of Friction

F(displacement)(change in inclination) = (1/2mv^2) + (mg)(change in height) + (mu)(Force Normal)(cosine theta)(displacement)

Change in Kinetic Energy = (1/2)mv^2

Change in Potential Energy = mg(change in height)

Work of Friction = (mu)(mg)(sin theta)

PE(0) + KE(0) = PE(1) + KE(1) + W(f)

Absolute Gravitational Potential Energy = F(G) = -(GM(e)m(2))/r

Power = Work / Time