A force F=(3 i+4 j) ;N acts on a particle moving in x-y plane. St… - Vidyadip

MCQ

A force $\vec{F}=(3 \vec{i}+4 \vec{j}) \;N$ acts on a particle moving in $x-y$ plane. Starting from origin, the particle first goes along $x$-axis to the point $(4,0) \,m$ and then parallel to the $y$-axis to the point $(4,3) \,m$. The total work done by the force on the particle is ............. $J$

A
$+12$
B
$-6$
C
$-12$
✓
$+24$

✓

Answer

Correct option: D.

$+24$

d
(d)

$\vec{F}=3 \hat{i}+4 \hat{j}$

Displacement vector $(\vec{x})=4 \hat{i}+3 \hat{j}$

$\vec{F} \cdot \vec{x}=(3 i+4 \hat{j}) \cdot(4 \hat{i}+3 \hat{j})$

$=12+12=24 \,J$

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