Question

a particle moves from a point $(-2\hat{i}+5\hat{j})$ to $(4\hat{i}+3\hat{k})$ when a force of $(4\hat{i}+3\hat{j})$ is applied . How much work has been done by the force?

Answer 1

Force = $4\hat{i}+3\hat{j}$

Displacement = $S_{final}-S_{initial}$

$=(4\hat{i}+3\hat{k})-(-2\hat{i}+\hat{j})$

$\Rightarrow 6\hat{i}-5\hat{j}+\hat{k}$

Work in x= $Fx.dx$

$=4\times 6$

$=24J$

Work in y= $Fy.dy$

$=3\times -5$

$=-15J$

work in z= $Fz.dz$

$=0\times 3=0J$

Work total = 24-15+0

= 9 J