Question

A variable force of\({\rm{5}}{{\rm{x}}^{{\rm{ - 2}}}}\)pounds move an object along a straight line when it is\({\rm{x}}\) feet from the origin. Calculate the work done in moving the object from\({\rm{x = 1}}\)ft to\({\rm{x = 10}}\)ft.

Short answer

The work done in moving the object is \(4.5{\rm{ }}lb \cdot ft\).

Step-by-step solution

Work done by the force

Work done by a force is defined as the product of the displacement of an object and the applied force which is in the direction of the object’s displacement.

If an object moves from a to b and \(f(x)\)is the force applied; work done is given as: \(W = \int_a^b f (x)dx\)

Given parameters

Applied force \(f(x) = 5{x^{ - 2}}\).

Displacement is from\(x = 1\) to\(x = 10\) i.e., \((a,b) = (1,10)\).

Calculating work done by force

Work done by the applied force is calculated as-

\(\begin{aligned}{}W = \int_a^b f (x)dx\\ = \int_1^{10} 5 {x^{ - 2}}dx\\ = \left( {\frac{{5{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right)_1^{10}\\ = \left( {\frac{{5{x^{ - 1}}}}{{ - 1}}} \right)_1^{10}\\ = \left( { - \frac{5}{x}} \right)_1^{10}\\ = \left( { - \frac{5}{{10}}} \right) - \left( { - \frac{5}{1}} \right)\\ = - 0.5 + 5\\ = 4.5{\rm{ }}lb \cdot ft\end{aligned}\)

Therefore, the work done in moving the object is \(4.5{\rm{ }}lb \cdot ft\).