## Precalculus (6th Edition) Blitzer

The work done by the crate is approximately $1115\text{ foot-pounds}$.
The amount of work done by a force $\mathbf{F}$ on an object from a point A to the point B is denoted by $W=\mathbf{F}\cdot \overrightarrow{AB}$. That is $W=\left\| \mathbf{F} \right\|\overrightarrow{\left\| AB \right\|}\cos \theta$ ; Where $\theta$ is the angle between the force and the direction of the motion. In the given case, the distance between the crate and the level of floor is 50 feet. \begin{align} & \left\| \overrightarrow{AB} \right\|=50 \\ & \theta =42{}^\circ \end{align} And, the magnitude of force $\left\| \mathbf{F} \right\|=30$. Therefore, the amount of work done will be calculated as below: \begin{align} & W=\left\| \mathbf{F} \right\|\overrightarrow{\left\| AB \right\|}\cos \theta \\ & =\left( 30 \right)\left( 50 \right)\cos 42{}^\circ \\ & =1500\left( 0.743144825 \right) \\ & =1114.71 \end{align} This implies $W\approx 1115$. So, work done by the crate is approximately $1115\text{ foot-pounds}$. Hence, the amount of work done by the crate is approximately $1115\text{ foot-pounds}$.