# Question #a7964

Oct 12, 2014

"When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. There are three key ingredients to work: force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement." Source:http://www.physicsclassroom.com/class/energy/Lesson-1/Definition-and-Mathematics-of-Work

The work equation is $\text{W}$ = $\text{F}$ x $d$, where W is work in Joules (J), F is force in N or ${\text{kg*m/s}}^{2}$, and $d$ is displacement. A Joule is a ${\text{kg*m"^2/"s}}^{2}$.

Known/Given:
Force, $\text{F}$ = $\text{137N}$
Work, $\text{W}$ = $\text{223J}$

Unknown:
displacement, $d$

Equation:
$\text{W}$ = $\text{F}$ x $d$

Solution: To solve for displacement, the work equation will have to be rearranged, so the $d$ = $\text{W"/"F}$. Then plug in your values.

$d$ = $\text{W"/"F}$ = $\text{223J"/"137N}$ = ${\text{223kg*m"^2/"s}}^{2}$$\div$ ${\text{137kg*m"/"s}}^{2}$ = $\text{1.63m}$ upward

When dividing a fraction by another fraction, you invert the second fraction and then multiply.

$d$ = $\text{W"/"F}$ = $\text{223J"/"137N}$ = ${\text{223kg*m"^2/"s}}^{2}$ x $\text{s"^2/ "137kg*m}$ = $\text{1.63m}$ upward

(The ${\text{s}}^{2}$ cancels, the $\text{kg}$cancels, and the $\text{m"^2/"m}$ reduces to $\text{m}$.)