Question

A force of $(10\hat{i}-3\hat{j}+6\hat{k})N$ acts on a body of 5$kg$ and displaces it from $A$ $(6\hat{i}+5\hat{j}-3\hat{k})m$ to $B(10\hat{i}-2\hat{j}+7\hat{k})m$ . The work done is

• A. zero
• B. $121J$
• C. $120J$
• D. none

Answer 1

The correct option is B $121J$

Work equals the dot product of force and displacement.

$W=F.d$

Force is given as :$10i-3j+6k$

Displacement is calculated from the final position minus the initial position.

$d=(10i-2j+7k)-(6i+5j-3k)=4i-7j+10k$

$W=(10i-3j+6k).(4i-7j+10k)$

$W=(10\times 4)+\left(\right(-3)\times (-7\left)\right)+(6\times 10)=40+21+60=121$

Hence, Option $B$ is correct.