Talk:PlanetPhysics/Basic Examples of Calculating Work in Physics

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\begin{document}

\textbf{Example 1}. Find in the various units (SI, English and CGS) the \htmladdnormallink{work}{http://planetphysics.us/encyclopedia/Work.html} done on a \htmladdnormallink{mass}{http://planetphysics.us/encyclopedia/Mass.html} 112 pounds when lifted through 100 feet.

For this example, the equation for work simplifies to just the \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html} times the distance that the force is acting.

$$ W = Fd $$

Plugging in the numbers for SI units

$$W = \frac{112 [lb]}{1} \cdot \frac{1 [slug]}{32.17 [lb]} \cdot \frac{14.59 [kg]}{1 [slug]} \times \frac{100 [ft]}{1} \frac{1 [m]}{3.281[ft]} \cdot 9.806 [m/s^2] = 15,180 [joules]$$

Plugging in the numbers for English units

$$ W = 112 [lb] \times 100 [ft] = 11,200 [ft-lb]$$

Plugging in the numbers for CGS units

$$ W = \frac{112 [lb]}{1} \cdot \frac{ 453.6 [gram]}{1 [lb]} \times \frac{100 [ft]}{1} \cdot \frac{ 1 [m]}{ 3.281 [ft]} \cdot \frac{100 [cm]}{ 1 [m]} \cdot \frac{9.806 [m/s^2]}{1} \cdot \frac{100 [cm]}{1 [m]} = 1.518 \times 10^{11} [ergs] $$

\textbf{Example 2}. How much mork is done by a force of $5x$ newtons acting in the x-direction upon a \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} while it is displaced from $x = 1 [m]$ to $x = 10 [m]$?

The force for this example is a \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} of $x$

$$ F(x) = 5x $$

integrating from \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html} 1 to 2 yields the work

$$ W = \int_1^{10} 5x dx = \bigg |_1^{10} \frac{5}{2} x^2 = 250 - 2.5 [N \cdot m] = 247.5 [N \cdot m] $$

\end{document}