Block B Velocity 44B2D8

1. **Problem Statement:**\nWe have a pulley system with two blocks: block A with mass $20\ \mathrm{kg}$ and block B with mass $14\ \mathrm{kg}$. Block A is released from rest. We need to find the velocity of block B after $2\ \mathrm{s}$, assuming upward is positive.\n\n2. **Understanding the system:**\nThe pulley system has one movable pulley and two fixed pulleys. The rope runs over the top pulley, down around the movable pulley, and up to an attachment point. Block B hangs from the movable pulley on the left side, and block A hangs lower on the right side.\n\n3. **Key formula and constraints:**\n- Let $a_A$ be the acceleration of block A downward (positive downward).\n- Let $a_B$ be the acceleration of block B upward (positive upward).\n- The rope length is constant, so the accelerations are related by the pulley system geometry.\n\n4. **Relation between accelerations:**\nBecause the movable pulley moves with block B, and the rope segments change length accordingly, the constraint is:\n$$2a_B = a_A$$\nThis means block A moves twice as fast downward as block B moves upward.\n\n5. **Apply Newton's second law:**\nFor block A (mass $m_A=20$):\n$$m_A g - T = m_A a_A$$\nFor block B (mass $m_B=14$):\n$$T - m_B g = m_B a_B$$\nwhere $T$ is the tension in the rope and $g=9.81\ \mathrm{m/s^2}$ is acceleration due to gravity.\n\n6. **Substitute $a_A = 2a_B$ into equations:**\nFrom block A:\n$$T = m_A g - m_A a_A = 20 \times 9.81 - 20 \times 2a_B = 196.2 - 40 a_B$$\nFrom block B:\n$$T = m_B g + m_B a_B = 14 \times 9.81 + 14 a_B = 137.34 + 14 a_B$$\n\n7. **Set tensions equal and solve for $a_B$: **\n$$196.2 - 40 a_B = 137.34 + 14 a_B$$\n$$196.2 - 137.34 = 14 a_B + 40 a_B$$\n$$58.86 = 54 a_B$$\n$$a_B = \frac{58.86}{54} = 1.09\ \mathrm{m/s^2}$$\n\n8. **Calculate velocity of block B after 2 seconds:**\nBlock B starts from rest, so velocity after time $t=2\ \mathrm{s}$ is:\n$$v_B = a_B t = 1.09 \times 2 = 2.18\ \mathrm{m/s}$$\n\n**Final answer:**\nThe velocity of block B after 2 seconds is $2.18\ \mathrm{m/s}$ upward.