A cord runs around two pulleys. A canister of mass 20 kg hangs from one pulley, and you exert a force on the free end of the cord. (a) What is the magnitude of the force if you lift the canister at constant speed? (b) To lift by 2 cm, how far must you pull the free end of the cord? During the lift, what is the work done on the canister by (c) your force and (d) the gravitational force?

a)

One of the pulley is fixed and the other is moving. From the moving pulley load of 20 kg is hanging. One end of the cord is fixed from ceiling and the other end is free. Let T be tension in the whole cord. T will be the force applied by me.

Since cord is wrapped around the moving pulley, 2T will balance the weight of load.

2T = mg

T = mg / 2

= 20 x 9.8 / 2

= 98 N.

b)

If weight moves upward by 2 cm, cord on either side of load becomes loose by 2 cm. So the free end of cord moves by 4 cm.

so free end has to be pulled by 4 cm.

c)

force applied by me = T

= 98 N

displacement = 4 cm

=.04 m

work done = force x displacement

= 98 x. 04

= 3.92 J.

d)

work done by gravitation

= mg x - d

= 20 x 9.8 x -.02

= - 3.92 J (work done will be negative as direction of displacement is opposite to direction of force by gravity).