In the figure, a cord runs around two massless, frictionless pulleys; a canister

Question

In the figure, a cord runs around two massless, frictionless pulleys; a canister with mass m = 50 kg hangs from one pulley; and you exert a force F on the free end of the cord. What must be the magnitude of F if you are to lift the canister at a constant speed?
2.45×102 N

To lift the canister by 4.1 cm, how far must you pull the free end of the cord?
8.20 cm

During that lift, what is the work done on the canister by your force (via the cord)?

(HELP!!)

During that lift, what is the work done on the canister by the weight mg of the canister? (Hint: When a cord loops around a pulley as shown, it pulls on the pulley with a net force that is twice the tension in the cord.)
-2.01×101 J

Explanation / Answer

Since Both the Force and displacement are downward, Therefore

Work done is given by = Force * Distance of free end of cord moved
W = 2.45 * 10^2 N * 8.2 *10^-2
Work done on the canister by your force = 20.09 J

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