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,
$\operatorname{Work}(\omega)= F \cdot s$
$=(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} $
$=-0.48 \,J$