Chapter 9, Problem 6T

To determine

The work done by the force $F=3i-5j$ in moving an object from the point $\left(2,2\right)$ to the point $\left(7,-3\right)$.

Answer to Problem 6T

The work done by the force $F=3i-5j$ in moving an object from the point $\left(2,2\right)$ to the point $\left(7,-3\right)$ is 90.

Explanation of Solution

Given:

The work done by the force $F=3i-5j$.

The two points are $\left(2,2\right)$ and $\left(7,-3\right)$.

Formula used: The vector v with initial point $P\left(x_1,y_1\right)$ and the terminal point $Q\left(x_2,y_2\right)$ is given below,

$v=\langle x_2-x_1,y_2-y_1\rangle$ (1)

The dot Product of the vectors $u=a_{1}i+a_{2}j$ and the vector $v=b_{1}i+b_{2}j$ is given below,

$u\cdot v=a_1{b}_1+a_2{b}_2$ (2)

The work W doe by a force F in moving along a vector D is given below,

$W=\text{F}\cdot \text{D}$ (3)

Calculation:

Substitute 2 for $x_1$, 2 for $y_1$, 7 for $x_2$ and $-3$ for $y_2$ in equation (1), to find the vector D passing through the points $\left(2,2\right)$ and $\left(7,-3\right)$,

$\begin{array}{c}\text{D}=\langle 7-2,-13-2\rangle \\ =\langle 5,-15\rangle \end{array}$

The force $F=3i-5j$ is written as $\text{F}=\langle 3,-5\rangle$.

Substitute 3 for $a_1$, $-5$ for $a_2$, 5 for $b_1$ and $-15$ for $b_2$ in equation (2), to find $\text{F}\cdot \text{D}$,

$\begin{array}{c}\text{F}\cdot \text{D=}\left(3\right)\left(5\right)+\left(-5\right)\left(-15\right)\\ =15+75\\ =90\end{array}$

Substitute 90 for $\text{F}\cdot \text{D}$ in equation (3), to find the work W,

$W=90$

Thus, the work done by the force $F=3i-5j$ in moving an object from the point $\left(2,2\right)$ to the point $\left(7,-3\right)$ is 90.