## Precalculus (6th Edition) Blitzer

Here, the force can be resolved into two vectors, such that: $\mathbf{F}=\left( 6\cos 40{}^\circ,6\sin 40{}^\circ \right)$ …… (1) As the object is moved from point $\left( 5,9 \right)$ to $\left( 8,20 \right)$, therefore, the point to which the object has traveled is: \begin{align} & \mathbf{AB}=\left[ \left( 8,20 \right)-\left( 5,9 \right) \right] \\ & =\left[ \left( 8-5 \right),\left( 20-9 \right) \right] \\ & =\left( 3,11 \right) \end{align} $\mathbf{AB}=3\mathbf{i}+11\mathbf{j}$ …… (2) Now, as it is known that: $\mathbf{W}=\mathbf{F}\cdot \mathbf{AB}$ …… (3) Substituting equation (1) and (2) in equation (3), \begin{align} & \mathbf{W}=\left( 6\cos 40{}^\circ,6\sin 40{}^\circ \right)\cdot \left( 3,11 \right) \\ & =\left[ \left( 6\cos 40{}^\circ \times 3 \right)+\left( 6\sin 40{}^\circ \times 11 \right) \right] \\ & =18\cos 40{}^\circ +66\sin 40{}^\circ \\ & \approx 56.21 \end{align}