Two Blocks On A Rough Incline

Question

A block of mass 5 kg lies on a frictionless incline of 37 degrees and is connected by a light string over a pulley at the top to a freely hanging 4 kg block. Taking g as 10 m/s^2 and sin 37 degrees as 0.6, find the acceleration of the system.

RankGuru · Accepted Answer

Consider both blocks as one system linked by an inextensible string over a frictionless pulley, so they share a common acceleration magnitude. The incline is frictionless, so the only driving and opposing agents are the gravity component of the incline block along the slope, $m_1 g\sin	heta$, and the full weight of the hanging block, $m_2 g$. The 5 kg block's pull down the slope is $5 	imes 10 	imes 0.6 = 30	ext{ N}$, exceeding the hanging block's weight $4 	imes 10 = 40	ext{ N}$? Here $m_2 g = 40	ext{ N}$ is larger, so the hanging block descends. Net force $= 40 - 30 = 10	ext{ N}$ acting on total mass $9	ext{ kg}$, giving $a = \frac{m_2 g - m_1 g\sin	heta}{m_1 + m_2} = \frac{40 - 30}{9} = 1.11	ext{ m/s}^2$. The 2.22 m/s^2 value halves the denominator mistakenly. The 0.55 m/s^2 value doubles it. The 0.22 m/s^2 value misuses cosine instead of sine when resolving the incline weight. Solving the two blocks separately would give a string tension of $T = m_2(g - a) = 4(10 - 1.11) \approx 35.6	ext{ N}$, comfortably between the hanging weight and zero, which validates the system result. This is the standard NCERT incline-pulley setup, where the frictionless surface means the only slope-direction force from the incline block is its gravity component. As a check, the small net 10 N over 9 kg gives a gentle acceleration, consistent with the near-balanced weights.

Two Blocks On A Rough Incline

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