Friction And Connected Blocks

Question

On a rough horizontal table a 4 kg block is pulled by a light string connected over a frictionless pulley to a hanging 2 kg block. If the coefficient of kinetic friction under the table block is 0.25 and g is 10 m/s^2, what is the acceleration of the system?

RankGuru · Accepted Answer

Treat the two blocks as one system sharing a single inextensible string, so they move with a common acceleration. The hanging 2 kg block's weight $m_2 g$ drives the motion, while kinetic friction $\mu m_1 g$ on the 4 kg table block opposes it. Applying Newton's Second Law to the system, $m_2 g - \mu m_1 g = (m_1 + m_2)a$. Substituting, $2(10) - 0.25 	imes 4 	imes 10 = (4 + 2)a$, giving $20 - 10 = 6a$, so $a = 10/6 = 1.67	ext{ m/s}^2$. The 3.33 m/s^2 value forgets friction and uses only the hanging weight. The 2.5 m/s^2 value omits the table block from the total mass. The 1.25 m/s^2 value overestimates the friction by misusing the normal force. Treating the blocks as a system conveniently sidesteps the internal tension, but one could equally solve the two Newton equations separately and find $T = m_2(g - a) = 2(10 - 1.67) \approx 16.7	ext{ N}$, a value between zero and the hanging weight as expected. This is the NCERT horizontal-pulley-with-friction problem, where the normal force on the table block is simply its own weight $m_1 g$ since the surface is horizontal. As a check, the result is positive yet below the frictionless 3.33 m/s^2, confirming that friction slows but does not halt the descent of the hanging mass.

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