As part a of the drawing shows, two blocks areconnected by a rope that passes ov
ID: 1738496 • Letter: A
Question
As part a of the drawing shows, two blocks areconnected by a rope that passes over a set of pulleys. One blockhas a weight of m1 = 512 N, and the other has a weight ofm2 = 708 N. The ropeand the pulleys are massless and there is no friction. (a) What is the acceleration of the lighter block? (b) Suppose that the heavier block is removed, and a downwardforce of 708 N is provided by someonepulling on the rope, as part b of the drawing shows. Findthe acceleration of the remaining block.As part a of the drawing shows, two blocks areconnected by a rope that passes over a set of pulleys. One blockhas a weight of m1 = 512 N, and the other has a weight ofm2 = 708 N. The ropeand the pulleys are massless and there is no friction. (a) What is the acceleration of the lighter block? (b) Suppose that the heavier block is removed, and a downwardforce of 708 N is provided by someonepulling on the rope, as part b of the drawing shows. Findthe acceleration of the remaining block.
Explanation / Answer
first we label: Fnet = F2-F1= mass of the system X accerationof the system: this is due to the heavier block pulling thelighter block down against the lighter blocks gravitationalforce. F1 = 512 N, F2 = 708. Mass 1 = F1/g, Mass 2 =F2/g, g = 9.8 m/s^2. once the masses are found the mass ofthe system can be calculated as (M1 + M2). And so: A (acceleration of the system and overall acceleration of thesmaller block) = (708N - 512N)/(M1 + M2). This accelerationshould be constant for any force = F2, so b.) should be thesame.
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