Two Blocks, One Rope, a Slipping Table
Block A (mass 4.0 kg) sits on a rough horizontal table (μ_k = 0.25). A light string runs over a frictionless pulley at the table edge and attaches to hanging Block B (mass 2.0 kg). The system is released from rest. A student draws a free-body diagram and writes 'T = m_B g' for block B, then uses that tension on block A. The string is taut throughout and does not stretch. Use g = 10 m/s².
1. Explain why T = m_B g is wrong here, and set up the correct equations for both blocks.
Read the worked answer
T = m_B g would hold only if B were in equilibrium (zero acceleration). Here B accelerates downward, so the net force on B is m_B g − T = m_B a. Setting T = m_B g forces a = 0 for B, which contradicts A also needing to accelerate.
Correct system (let a = magnitude of acceleration, same for both since string is inextensible):
- Block A (horizontal): T − f_k = m_A a, with f_k = μ_k m_A g = 0.25(4.0)(10) = 10 N.
- Block B (downward positive): m_B g − T = m_B a → 20 − T = 2.0a.
2. Solve for the acceleration and tension, then check the tension against the student's value.
Read the worked answer
Add equations to eliminate T: (T − 10) + (20 − T) = 4.0a + 2.0a → 10 = 6.0a → a = 1.67 m/s².
T from A: T = 10 + 4.0(1.67) = 16.7 N. Check with B: T = 20 − 2.0(1.67) = 16.7 N ✓.
Student's value T = m_B g = 20 N is too large by ~3.3 N. The tension is less than the hanging weight because B accelerates downward - a classic internal-consistency check.
3. Suppose μ_k were large enough that the blocks don't move. What would T be, and does the student's formula become valid?
Read the worked answer
If the system stays at rest, a = 0 for both. Then for B: T = m_B g = 20 N. For A: static friction balances T, requiring f_s = 20 N ≤ μ_s m_A g. So yes - only in the static case does T = m_B g hold. This is the judgment: the student's formula is a special case (equilibrium), not a general law. Recognizing when a shortcut is licensed is the skill being tested, not memorizing T = mg.