Question

A force $\overrightarrow{F}=6\hat{i}-8\hat{j}N$, acts on a particle and displaces it over $4m$ along the positive $X$-axis and then displaces it over $6m$ along the positive $Y$-axis. The total work done over the two displacements is:

• A. $72J$
• B. $24J$
• C. $-24J$
• D. $zero$

Answer 1

The correct option is C $-24J$

$W.D=\int \overrightarrow{F}.\overrightarrow{ds}$
$=(6\hat{i}-8\hat{j}).\left(4\hat{i}\right)+(6\hat{i}-8\hat{j}).\left(6\hat{j}\right)$
$=(6\times 4)+(-8\times 6)=24-48=-24J$