Question

How much work is done by a force vector F = (4x N)i hat +(5 N)j, with x in meters, that moves a particle from a position vector r i = (5 m)i hat +(6 m)j, to a position vector r f = -(2 m)i hat -(5 m)j?

Answer 1

Answer:

- 83 J

Explanation:

$\overrightarrow{F}=4\widehat{i}+5\widehat{j}$

$\overrightarrow{r_{i}}=5\widehat{i}+6\widehat{j}$

$\overrightarrow{r_{f}}=-2\widehat{i}-5\widehat{j}$

$\overrightarrow{d}=\overrightarrow{r_{f}}-\overrightarrow{r_{i}}$

$\overrightarrow{d}=\left ( -2-5 \right )\widehat{i}+\left ( -5-6 \right )\widehat{j}$

$\overrightarrow{d}=-7\widehat{i}-11\widehat{j}$

Work done

$W=\overrightarrow{F}.\overrightarrow{d}$

$W=\left (4\widehat{i}+5\widehat{j}  \right ).\left (-7\widehat{i}-11\widehat{j}  \right )$

W = - 28 - 55

W = - 83 J

Thus, the work done is - 83 J.