Chapter 9, Problem 114RE

To determine

To find: The work done by a force of 5 pounds acting in the direction $60°$ to the horizontal.

Answer to Problem 114RE

The work done by the force of $5\text{Lb}$ is $50\text{ft}$ .

Explanation of Solution

Given:

The distance covered is $20\text{feet}$ .

The initial coordinates are $\left(0,0\right)$ to $\left(20,0\right)$ .

Calculation:

Consider the diagram for the given case is shown in Figure 1

Figure 1

Here, the point $A=\left(0,0\right)$ and point $B$ are $\left(20,0\right)$ . Then the vector $\overrightarrow{AB}=20\text{i}$ .

Consider the angle between the vector $v$ and the $i$ is $\alpha$ , than vector is expressed as,

Then, the force vector is obtained as,

  $\begin{array}{c}\left|F\right|=\left|\left|F\right|\right|\left(\cos \alpha i+j\sin \alpha \right)\\ =\left|\left|F\right|\right|\left(\cos 60°i+j\sin 60°\right)\\ =\frac{5}{2}i+\frac{5\sqrt{3}}{2}j\end{array}$

Consider the work done by the force is,

  $W=F\cdot \overrightarrow{AB}$

Then,

  $\begin{array}{c}W=\left(\frac{5}{2}i+\frac{5\sqrt{3}}{2}j\right)\cdot \left(20i+0j\right)\\ =\frac{5}{2}\cdot 20+\frac{5\sqrt{3}}{2}\\ =50\end{array}$

Thus, the work done by the force of $5\text{Lb}$ is $50\text{ft}$ .