Two identical an ideal pulley masses, 25.0kg each, are through an ideal string a

Question

Two identical an ideal pulley masses, 25.0kg each, are through an ideal string and as shown. The angles are theta = 30.0 degree, and gamma = 60.0 degree. (I) Draw one free body diagram for each of the two masses, with the positive horizontal direction corresponding to the direction of motion. (II) The magnitude of the acceleration of the blocks is A. 0.60 m/s^2. B. 3.47 m/s^2 C. 1.96 m/s^2. D. 1.80 m/s^2 E. 2.37 m/s^2 (III) If the blocks start from rest, the time required to move a distance of 98.0cm is A. 0.752s. B. 1.42s C. 1.04s D. 0.909s E. 1.80s.

Explanation / Answer

let the tension in the string = T

let's devlop equations of motion for both masses

25 (a) = 25 (9.8) sin 60 - T

25 a= 212.176 - T -----------eq(1)

25(a) = T - 25 (9.8) sin 30

25(a) = T - 122.5 ----------eq(2)

adding eq (1) , and (2)

50 a= 89.676

a= 1.79 m/s^2 apprx ( option D)

b) s = 1/2 at^2

0.98 = 0.5 ( 1.80) t^2

t = 1.04 sec ( option C)

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