(c7p63)In the figure, a cord runs around two massless, frictionless pulleys; a c

Question

(c7p63)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. a)What must be the magnitude of F if you are to lift the canister at a constant speed?b)To lift the canister by 2.6 cm, 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)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.) with explanation steps if possible.

Explanation / Answer

solution:

note: the diagram link you provided is asking username and password.

  I am visualising what the diagram is like, I am assuming the cord is fixed overhead 1 end, passes around 1 pulley to which the cannister is attached, up and over a second overhead pulley and down to where the effort is applied.
The MA of such a system will be 2:1.
1) F = (50 x 9.8)/2 = 245N.
2) (2.6 x 2) = 5.2cm. =.052
3) Work = (force applied x distance load moves), = 245 x.052 = 12.74 joules.
4) mg = 490N. Distance it moves = 6cm. (.06m). work done by gravitational force(or mg) =(490 x .026) = 12.74 Joules

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