How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.33? Assume no friction acts on the wagon.
Solution:
The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,
W=Fd \cos \theta
In this case, we are given the following values:
\begin{align*}
F & = 50\ \text{N} \\
d & = 30\ \text{m} \\
\theta & = 30^{\circ}
\end{align*}Substituting these values into the equation, we have
\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 50\ \text{N} \right)\left( 30\ \text{m} \right) \cos 30^{\circ } \\
W & = 1299.0381\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}