How to calculate the work done by a variable force F(x)
The formula we use to calculate the work done when a variable force is applied to an object
To calculate the work done when a variable force is applied to lift an object of some mass or weight, we’ll use the formula
???W=\int^b_aF(x)\ dx???
where ???W??? is the work done, ???F(x)??? is the equation of the variable force, and ???[a,b]??? is the starting and ending height of the object.
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If ???W??? is positive, it means that the force is doing work in the given interval. If ???W??? is negative, then work needs to be done on the interval. The answer to these types of work problems is usually given in Joules J.
How to calculate the work done by some variable force (when we’re given a force function F(x))
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Finding the work done, given a variable force F(x)
Example
Find the work done to lift a ???20??? kg box from the floor to a height of ???3??? m when the variable force ???F(x)??? is given in Newtons.
???F(x)=4x^2-2x+3???
Using the formula from this section, and defining the interval ???[a,b]??? as ???[0,3]???, we get
???W=\int^3_04x^2-2x+3\ dx???
???W=\int^3_04x^2\ dx+\int^3_0-2x\ dx+\int^3_03\ dx???
???W=4\int^3_0x^2\ dx-2\int^3_0x\ dx+3\int^3_01\ dx???
If W is positive, it means that the force is doing work in the given interval. If W is negative, then work needs to be done on the interval.
Integrating, we get
???W=\left[4\left(\frac{x^3}{3}\right)-2\left(\frac{x^2}{2}\right)+3(x)\right]\Bigg|^3_0???
???W=\frac{4x^3}{3}-x^2+3x\Big|^3_0???
Now we’ll evaluate over the interval.
???W=\left(\frac{4(3)^3}{3}-(3)^2+3(3)\right)-\left(\frac{4(0)^3}{3}-(0)^2+3(0)\right)???
???W=36???
???36??? J of force are required to lift a ???20??? kg box from the floor to a height of ???3??? m when the variable force applied is defined by ???F(x)=4x^2-2x+3???.
