Question

Shown is the graph of a force function (in newtons) thatincreases to itsmaximum value and then remains constant.How much work is done by theforce in moving an object adistance of 8 m?

Short answer

The work done by the force in moving an object a distance of \(8m\)is\(180{\rm{ }}joules\).

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 \(f(x)\)can be written as-

For \(0 \le x \le 4,{\rm{ }}f(x) = \frac{{30x}}{4} = \frac{{15x}}{4}\)

For \(4 \le x \le 8,{\rm{ }}f(x) = 30\)

\(f(x) = \left\{ {\begin{aligned}{{}}{\frac{{15x}}{2}{\rm{ }}if{\rm{ }}0 \le x \le 4}\\{30{\rm{ }}if{\rm{ }}4 \le x \le 8}\end{aligned}} \right.\)

Calculating work done by force

Work done by the applied force is calculated as-

\(\begin{aligned}W &= \int_a^b f (x)dx\\ &= \int_0^8 f (x)dx\\ &= \int_0^4 f (x)dx + \int_4^8 f (x)dx\\ &= \int_0^4 {\frac{{15}}{2}} xdx + \int_4^8 3 0dx\\ &= \left( {\frac{{15}}{2} \cdot \frac{{{x^2}}}{2}} \right)_0^4 + (30x)_4^8\\ &= \left( {\frac{{15}}{4} \cdot {4^2} - 0} \right) + (30(8) - 30(4))\\ &= \left( {15 \cdot 4} \right) + (240 - 120)\\ &= 60 + 120\\ &= 180\end{aligned}\)

Therefore, the work done by the force in moving an object a distance of \(8m\)is \(180{\rm{ }}joules\).