Newton's Laws of Motion: Step-by-Step Worked Examples (6)

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Question

A block of mass m1=4m_1 = 4 kg sits on a frictionless table connected by a light inextensible string over a frictionless pulley to a hanging block of mass m2=6m_2 = 6 kg. Find the acceleration of the system and the tension in the string. Take g=10m/s2g = 10\,\text{m/s}^2.

This appeared in a JEE Main 2024 PYQ variant and tests whether we can write FBDs cleanly without skipping signs.

Solution — Step by Step

For m1m_1 on the table, the string tension TT pulls it horizontally toward the pulley. For m2m_2 hanging, gravity m2gm_2 g pulls down and tension TT pulls up. Pick the direction of motion as positive — we expect m2m_2 to fall, so down is positive for it and the table block accelerates toward the pulley.

For m1m_1: T=m1aT = m_1 a. For m2m_2: m2gT=m2am_2 g - T = m_2 a. The string is inextensible, so both blocks share the same magnitude aa.

Adding gives m2g=(m1+m2)am_2 g = (m_1 + m_2)a, so

a=m2gm1+m2=6×1010=6m/s2a = \frac{m_2 g}{m_1 + m_2} = \frac{6 \times 10}{10} = 6\,\text{m/s}^2

T=m1a=4×6=24NT = m_1 a = 4 \times 6 = 24\,\text{N}. Quick sanity check: T<m2g=60T < m_2 g = 60 N, which must be true otherwise m2m_2 would not fall.

Final answer: a=6m/s2a = 6\,\text{m/s}^2, T=24NT = 24\,\text{N}.

Why This Works

The string constraint is the key idea. Because the string does not stretch, the two blocks have the same speed at every instant, hence the same acceleration magnitude. Newton’s third law means the tension pulling m1m_1 forward is the same magnitude as the tension pulling m2m_2 up — that is what lets us call both tensions TT.

Once we accept these two ideas, the pulley problem reduces to two scalar equations in two unknowns. The trick of “adding the equations” works because TT appears with opposite signs in the two FBDs, so it cancels.

Alternative Method

We can treat the system as one composite object. The only external force along the direction of motion is m2gm_2 g (the table’s normal force and m1gm_1 g are perpendicular to motion). The total mass being accelerated is m1+m2m_1 + m_2. So a=m2g/(m1+m2)=6m/s2a = m_2 g / (m_1 + m_2) = 6\,\text{m/s}^2 directly. Use this shortcut only after you can do the FBD method — JEE often gives a twist (friction, incline) where the shortcut breaks.

A common mistake is writing Tm2g=m2aT - m_2 g = m_2 a for the hanging block. That assumes the block moves up. Always pick a sign convention first, then stick to it. If aa comes out negative, the block moves the other way — no harm done.

Common Mistake

Students often forget that for m1m_1 on a smooth horizontal table, gravity and normal force cancel, so the only horizontal force is TT. Writing m1gT=m1am_1 g - T = m_1 a here is wrong — gravity does not act along the direction of motion. Always resolve forces along the direction of motion before applying F=maF = ma.

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