Discussion Forum

Consider  a particle  on which  constant forces $F_{1}=\hat{i}+2\hat{j}+3\hat{k}$ N and $F_{2}=4\hat{i}-5\hat{j}-2\hat{k}$ N act together resulting in a displacement from position  $r_{1}=20\hat{i}+15\hat{j}$ cm to $r_{2}=7\hat{k}$  cm. The  total work done on the particle is 

Answer:

Explanation:

Given, 

$F_{1}=\hat{i}+2\hat{j}+3\hat{k}$ N

$F_{2}=4\hat{i}-5\hat{j}-2\hat{k}$ N

$r_{1}=20\hat{i}+15\hat{j}$ cm

$r_{2}=7\hat{k}$  cm

Total  force on particle , $F= F_{1}+F_{2}$

               =$\hat{i}+2\hat{j}+3\hat{k}+4\hat{i}-5\hat{j}-2\hat{k}$

= $(5 \hat{i}-3\hat{j}+\hat{k})N$

 Displacement of pratice , $s=r_{2}-r_{1}$

= $7 \hat{k}-20\hat{i}-15\hat{j}$

  =$(-20 \hat{i}-15 \hat{j}+7\hat{k})$cm

 We know that,

Work (w)=F.s

 = $(5\hat{i}-3\hat{j}+\hat{k})(-20 \hat{i}-15 \hat{j}+7\hat{k}) \times 10^{-2}$

  =$(-100+45+7) \times 10^{-2}$

 =-0.48 J