Question

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

• A. -0.48 J
• B. + 0.48 J
• C. - 4.8 J
• D. + 4.8 J

Answer 1

Answer: (A)

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 particle, $s=r_2-r_1$
$=7\hat{k}-20\hat{i}-15\hat{j}$
$=(-20\hat{i}-15\hat{j}+7\hat{k})cm$
We know that,
$\mathrm{Work}(\omega )=F\cdot s$
$=(5\hat{i}-3\hat{j}+\hat{k})(-20\hat{i}-15\hat{j}+7\hat{k})\times 10^{-2}$
$=(-100+45+7)\times 10^{-2}$
$=-0.48J$