FIGURE 6-13
-Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses?

Question

FIGURE 6-13 FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses? A) 0.981 m/s2 B) 1.97 m/s2 C) 2.74 m/s2 D) 2.26 m/s2 E) 0 m/s2

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses? A) 0.981 m/s<sup>2</sup> B) 1.97 m/s<sup>2</sup> C) 2.74 m/s<sup>2</sup> D) 2.26 m/s<sup>2</sup> E) 0 m/s<sup>2</sup>

-Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses?

Quizplus · Accepted Answer

The tension in the string above block A is T1 = (6.00 kg)(9.81 m/s^2) = 58.86 N. The tension in the string above block C is T3 = (3.00 kg)(9.81 m/s^2) = 29.43 N. The net force on block A is (58.86 N) - (T2) = (6.00 kg)(a), where T2 is the tension in the string between blocks A and B. The net force on block C is T3 - (4.00 kg)(a). Since the pulleys are ideal, the tension in the string between blocks A and B is equal to the tension in the string between blocks B and C, so T2 = T3. Solving the system of equations, we find a = 2.26 m/s^2.

FIGURE 6-13 -Refer to Figure 6-13. Block a Has a Mass of a Mass