The work done by a force \( \vec{F} \) during a displacement \( \vec{r} \) is given by \( \vec{F} \cdot \vec{r} \). Suppose a force of \( 12 \mathrm{~N} \) acts on a particle in vertically upward direction and the particle is displaced through $2.0m$ in vertically downward direction. Find the work done by the force during this displacement.
Given,
Force (F) = 12N
Displacement (s) = 2.0m
To find
Work Done (W) = ?
We know that,
$W=F\times s\times cos\Theta $
where, W = Work done, F = Force, s = displacement, and $\Theta $= angle between the force and the direction of motion.
Substituting the value into the formula-
$W=F\times s\times cos\Theta $
$W=12\times 2.0\times cos0\unicode{xb0} $
$W=24\times 1$ $[\because cos0\unicode{xb0} =1]$
$W=24 Joule$
Hence, the work done by the force during this displacement is $24 J$
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