FIGURE 6-11
-Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses?

Question

FIGURE 6-11 FIGURE 6-11 -Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses? A) 3.22 m/s2 B) 5.10 m/s2 C) 3.92 m/s2 D) 6.54 m/s2 E) 8.24 m/s2

FIGURE 6-11 -Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses? A) 3.22 m/s<sup>2</sup> B) 5.10 m/s<sup>2</sup> C) 3.92 m/s<sup>2</sup> D) 6.54 m/s<sup>2</sup> E) 8.24 m/s<sup>2</sup>

-Refer to Figure 6-11. Block A has a mass of 3.00 kg and rests on a smooth table and is connected to block B, which has a mass of 2.00 kg, after passing over an ideal pulley, as shown. Block B is released from rest. What is the acceleration of the masses?

Quizplus · Accepted Answer

Since there is no friction between block A and the table, the tension in the string is equal on both sides of the pulley. Applying Newton's second law to block A, we get $\sum F = ma_A = T$. For block B, we have $\sum F = ma_B = mg - T$, where $g$ is the acceleration due to gravity. Since both blocks have the same acceleration, we can eliminate $T$ between these two equations to get $a = g\frac{m_A - m_B}{m_A + m_B}$. Plugging in the given values, we get $a = 9.81 \frac{3.00-2.00}{3.00+2.00} \approx 3.92 	ext{ m/s}^2$. Therefore, the correct answer is choice C.

FIGURE 6-11 -Refer to Figure 6-11. Block a Has a Mass of a Mass