A) In the figure, a cord runs around two massless, frictionless pulleys; a canis
ID: 2178398 • Letter: A
Question
B)To lift the canister by 2.6cm, 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.)
Explanation / Answer
a)F = T
mg = 2T
F = T = mg/2 = 220.725 N
b) x = 2 x 2.6 = 5.2 cm
c) W = F.d = 220.725 x (5.2/100) = 11.48 J
d) W = - mg x 2.6/100 = 45 x 9.18 x 2.6/100 = - 11.48 J