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Q. A person holds a bucket of weight $60 \, N$ . He walks $7 \, m$ along the horizontal path and then climbs up a vertical distance of $5 \, m.$ The work done by the man is

NTA AbhyasNTA Abhyas 2022

Solution:

No work is done while covering the horizontal distance because $\overset{ \rightarrow }{F}.\overset{ \rightarrow }{s}=0 \, \left(\because \theta = 90 ^\circ \right)$
But work is done during vertical displacement which is given by
$Fh=60\times 5=300 \, J$