Chapter 9.2, Problem 47E

To determine

To find: The work done by a constant force in moving a lawn mower at a distance of 200ft along a horizontal path.

Answer to Problem 47E

The work done by a constant force in moving a lawn mower is 8660.25 ft-lb.

Explanation of Solution

Given:

The distance pushed is 200ft.

A constant force along a horizontal path is 50lb.

The angle formed between the handle of lawn mower and horizontal axis is $30°$.

Formula used:

The work done (W) in moving an object along a vector D by a force F is,

$W=F\cdot D$ (1)

Calculation:

The lawn mower moves 200ft towards left along the x-axis and there is no change in y-axis. So, the x-component is 200ft and y-component is 0.

The displacement vector is,

$D=-200i$

The formula to calculate force with given length and direction in terms of components is,

$F=\left|F\right|\cos \theta i+\left|F\right|\cos \theta j$

Substitute 50 for $\left|F\right|$ and $30°$ for $\theta$ in the above formula to find force.

$\begin{array}{c}F=-\left(50\cos 30°\right)i-\left(50\sin 30°\right)j\\ \text{=}-\left(50\times \frac{\sqrt{3}}{2}\right)i-\left(50\times \frac{1}{2}\right)j\\ =-25\sqrt{3}i-25j\end{array}$

Substitute $-25\sqrt{3}i-25j$ for F and $-200i$ for D in equation (1) to find work done.

$\begin{array}{c}W=\left(-25\sqrt{3}i-25j\right)\cdot \left(-200i\right)\\ =\left(-25\sqrt{3}\right)\times \left(-200\right)-\left(-25\right)\times 0\\ =8660.25\end{array}$

Thus, the work done by a constant force in moving a lawn mower is 8660.25 ft-lb.