Question

If two forces $F_1=2\hat{i}+4\hat{k}$ $F_2=3\hat{j}+2\hat{k}$ acts on one body and displaces it from (1,0,0) to (2,1,1) find net work done.

Answer 1

Given,

$f_1=2\hat{i}+4\hat{k}$

$f_2=3\hat{j}+2\hat{k}$

So the displacement of the particle is:

$s=b-a=(2-1)\hat{i}+(1-0)\hat{j}+(1-0)\hat{k}$

$s=2\hat{i}+1\hat{j}+1\hat{k}$

$f=f_1+f_2=(2+0)\hat{i}+(0+3)\hat{j}+(4+2)\hat{k}$

$f=2\hat{i}+3\hat{j}+6\hat{k}$

So the work done is $f\times s$

$w=2\hat{i}+3\hat{j}+6\hat{k}\times 2\hat{i}+1\hat{j}+1\hat{k}$

$4+3+6=13Nm$