Answer:
[tex]2.70J[/tex].
Step-by-step explanation:
We have been given that a variable force of [tex]3x^{-2}[/tex] pounds moves an object along a straight line when it is x feet from the origin. We are asked to find the work done in moving the object from x = 1 ft to x = 10 ft.
We know that [tex]dW=Fdx[/tex], where F represents force.
[tex]dW=3x^{-2}dx[/tex]
Now, we will integrate both sides of equation as:
[tex]\int\limits{dw} =\int\limits^{10}_1 {3x^{-2} \, dx[/tex]
Take the constant out:
[tex]\int\limits{1dw} =3\int\limits^{10}_1 {x^{-2} \, dx[/tex]
[tex]w =3\int\limits^{10}_1 {x^{-2} \, dx[/tex]
Upon applying power rule of integrals, we will get:
[tex]w =3\/[ \frac{x^{-2+1}}{-2+1} ]^{10}_1[/tex]
[tex]w =3[ \frac{x^{-1}}{-1} ]^{10}_1[/tex]
[tex]w =3[ -\frac{1}{x} ]^{10}_1[/tex]
[tex]w =3[ -\frac{1}{10} -(-\frac{1}{1})][/tex]
[tex]w =3[ -\frac{1}{10}+\frac{10}{10})][/tex]
[tex]w =3[\frac{9}{10}][/tex]
[tex]w=2.70[/tex]
Therefore, the work done by the object is [tex]2.70J[/tex].
