Question

Calculate the amount of work done by a boy
when
(i) the boy holds the bundle of books of mass
5 kg for 5 minutes
the boy walks with the same bundle of
books along a level road at a speed of 5
ms!
(iii) the boy lifts up the same bundle of books
by 1 m in order to kept it on a book shelf,
(g = 9.8 ms?).


plz answer it on a paper​

Answer 1

Answer:

In case 1, displacement is zero , hence the work done by the boy is zero.

In case 2, the force applied ( which is equal to weight of books) is perpendicular to the displacement. hence

we know , work done = F.S

w= Fs cos(theta)

w=Fs cos90°

w=0 (cos90°= 0)

In case 3 , work done , W= F.s =Fs cos180°( force and displacement are in opposite direction)

w= mg×s×cos0° ( F=mg)

w= 5×9.8×1× -1 (cos180°= -1)

w= -49 joule.