1as it is slowly slid up
at any moment the block should be in equillibrium
in the first case force along the incline in downward direction = mgsin37 =3mg/5
so the force applied by the person =3mg/5
F.ds= 3mg/5*2 =6mg/5= 6*100/5 = 120 J
in the second case there are two components of the man's force
one perpendicular to the incline and one paralell to it
the work done by the part perpendicular to the incline is zero as there is no displace ment of the block in that direction
but as the block remains in equillibrium always
the component of man's force along the incline should be again = 3mg/5
and hence the same answer follows
