Q. A person holds a bucket of weight 60 N. He walks 7m along the horizontal path and then climbs up a vertical distance of 5 m. The work done by the man is
A) 300 J
B) 420 J
C) 720 J
D) None of these
Explanation
(a) No work is done while covering the horizontal distance because $\overrightarrow{\mathbf{F}}.\overrightarrow{\mathbf{s}} = 0(\therefore\ \theta = 90{^\circ})$
But work is done during vertical displacement which is given by
$Fh = 60 \times 5 = 300\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.