In the figure below, a cord runs around two massless, frictionless pulleys. A ca
ID: 2202293 • Letter: I
Question
In the figure below, a cord runs around two massless, frictionless pulleys. A canister with mass m = 11 kg hangs from one pulley, and you exert a force on the free end of the cord. (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.)
(b) To lift the canister by2.5cm, how far must you pull the free end of the cord?
(c) During that lift, what is the work done on the canister by your force (via the cord)?
(d) What is the work done on the canister by the gravitational force on the canister?
Explanation / Answer
( a ) Pulley 1 : v = cte
---> Fnet =0
---> 2T - mg = 0
--> T = mg/2 = 11*9.8/2
--> T = 53.9 N
Hand - cord : T - F = 0
--> F =mg/2
---> F = 53.9 N ---> Answer
(b) To rise “m” 0.025m, two segments of the cord must be shorten by that amount. Thus, the amount of the string pulled down at the left end is: 0.05m---> Answer
(c ) WF = F * d = (53.9 N )(0.05 m ) = 2.695 J---> Answer
( d ) WFg = - mgd =- (0.025 m )( 11 kg )(9.8m / s^ 2 ) = - 2.695 J---> Answer
WF+WFg=0 There is no change in kinetic energy.
This is consistent since There is no change in kinetic energy.