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Example 1. Find in the various units (SI, English and CGS) the work done on a mass 112 pounds when lifted through 100 feet.
For this example, the equation for work simplifies to just the force times the distance that the force is acting.
Plugging in the numbers for SI units
Plugging in the numbers for English units
Plugging in the numbers for CGS units
Example 2. How much mork is done by a force of  newtons acting in the x-direction upon a particle while it is displaced from ![$x = 1 [m]$](http://images.physicslibrary.org/cache/objects/826/l2h/img6.png "$x = 1 [m]$") to
![$x = 10 [m]$](http://images.physicslibrary.org/cache/objects/826/l2h/img7.png "$x = 10 [m]$") ?
The force for this example is a function of 
integrating from position 1 to 2 yields the work
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![[parent]](https://images.physicslibrary.org/images/uparrow.png)
![$\displaystyle W = \frac{112 [lb]}{1} \cdot \frac{1 [slug]}{32.17 [lb]} \cdot \f... ...frac{100 [ft]}{1} \frac{1 [m]}{3.281[ft]} \cdot 9.806 [m/s^2] = 15,180 [joules]$](http://images.physicslibrary.org/cache/objects/826/l2h/img2.png "$\displaystyle W = \frac{112 [lb]}{1} \cdot \frac{1 [slug]}{32.17 [lb]} \cdot \f... ...frac{100 [ft]}{1} \frac{1 [m]}{3.281[ft]} \cdot 9.806 [m/s^2] = 15,180 [joules]$")
![$\displaystyle W = 112 [lb] \times 100 [ft] = 11,200 [ft-lb]$](http://images.physicslibrary.org/cache/objects/826/l2h/img3.png "$\displaystyle W = 112 [lb] \times 100 [ft] = 11,200 [ft-lb]$")
![$\displaystyle W = \frac{112 [lb]}{1} \cdot \frac{ 453.6 [gram]}{1 [lb]} \times ... ...c{9.806 [m/s^2]}{1} \cdot \frac{100 [cm]}{1 [m]} = 1.518 \times 10^{11} [ergs] $](http://images.physicslibrary.org/cache/objects/826/l2h/img4.png "$\displaystyle W = \frac{112 [lb]}{1} \cdot \frac{ 453.6 [gram]}{1 [lb]} \times ... ...c{9.806 [m/s^2]}{1} \cdot \frac{100 [cm]}{1 [m]} = 1.518 \times 10^{11} [ergs] $")
![$\displaystyle W = \int_1^{10} 5x dx = \bigg \vert _1^{10} \frac{5}{2} x^2 = 250 - 2.5 [N \cdot m] = 247.5 [N \cdot m] $](http://images.physicslibrary.org/cache/objects/826/l2h/img10.png "$\displaystyle W = \int_1^{10} 5x dx = \bigg \vert _1^{10} \frac{5}{2} x^2 = 250 - 2.5 [N \cdot m] = 247.5 [N \cdot m] $")