FIGURE 6-13
-Refer to Figure 6-13. Block A has a mass of 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 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 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 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) 1.86 m/s2 C) 2.94 m/s2 D) 4.20 m/s2 E) 4.71 m/s2

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 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) 1.86 m/s<sup>2</sup> C) 2.94 m/s<sup>2</sup> D) 4.20 m/s<sup>2</sup> E) 4.71 m/s<sup>2</sup>

-Refer to Figure 6-13. Block A has a mass of 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 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 acceleration of the masses is given by:$$a = \frac{F_{net}}{m_{total}}$$where $F_{net}$ is the net force acting on the system and $m_{total}$ is the total mass of the system.In this case, the net force acting on the system is:$$F_{net} = T - f_g$$where $T$ is the tension in the rope and $f_g$ is the force of gravity acting on the system.The tension in the rope is:$$T = m_Bg$$where $m_B$ is the mass of block B and $g$ is the acceleration due to gravity.The force of gravity acting on the system is:$$f_g = (m_A + m_B + m_C)g$$where $m_A$ is the mass of block A, $m_B$ is the mass of block B, and $m_C$ is the mass of block C.Substituting these expressions into the equation for the acceleration, we get:$$a = \frac{m_Bg - (m_A + m_B + m_C)g}{m_A + m_B + m_C}$$$$a = \frac{(5.00 	ext{ kg})(9.81 	ext{ m/s}^2) - (3.00 	ext{ kg} + 5.00 	ext{ kg} + 2.00 	ext{ kg})(9.81 	ext{ m/s}^2)}{3.00 	ext{ kg} + 5.00 	ext{ kg} + 2.00 	ext{ kg}}$$$$a = 0.981 	ext{ m/s}^2$$Therefore, the acceleration of the masses is 0.981 m/s^2.

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