Newton's Laws of Motion: Speed-Solving Techniques (10)

easy 3 min read

Question

Two blocks of masses 3kg3 \, \text{kg} and 5kg5 \, \text{kg} are connected by a light string and pulled along a smooth horizontal floor by a force F=32NF = 32 \, \text{N} applied to the 5kg5 \, \text{kg} block. Find the acceleration of the system and the tension in the string — in under 30 seconds.

Solution — Step by Step

When two blocks move together with the same acceleration, we can lump their masses. Total mass =3+5=8kg= 3 + 5 = 8 \, \text{kg}.

Acceleration a=F/Mtotal=32/8=4m/s2a = F / M_{\text{total}} = 32 / 8 = 4 \, \text{m/s}^2.

Pick the simpler block — the 3kg3 \, \text{kg} one. The only horizontal force on it is the tension TT.

T=m1a=3×4=12NT = m_1 \cdot a = 3 \times 4 = 12 \, \text{N}.

For the 5kg5 \, \text{kg} block: net force =FT=3212=20N= F - T = 32 - 12 = 20 \, \text{N}. Then a=20/5=4m/s2a = 20 / 5 = 4 \, \text{m/s}^2. Matches.

Final answers: a=4m/s2a = 4 \, \text{m/s}^2, T=12NT = 12 \, \text{N}.

Why This Works

The “system + isolation” combo is the fastest way to crack connected-body problems. The system trick gives us acceleration in one line. The isolation trick then gives any internal force we need.

We never write the messy two-equation, two-unknown system. We never substitute. The problem collapses into two divisions.

30-second rule: For any connected-block problem with one external force, the tension on the “rear” block always equals (rear mass / total mass) × applied force. Here: (3/8)×32=12N(3/8) \times 32 = 12 \, \text{N}. We can write the answer instantly.

Alternative Method — Free Body Diagrams

The textbook approach: write FT=m2aF - T = m_2 a and T=m1aT = m_1 a. Add the two equations to get F=(m1+m2)aF = (m_1 + m_2)a, solve for aa, then back-substitute. Same answer, twice the work.

For JEE Main where every second counts, the system trick is non-negotiable. We save 60-90 seconds per question, which compounds across the paper.

Common Mistake

Students apply a=F/m1a = F/m_1 or a=F/m2a = F/m_2 instead of total mass. This happens when we forget that both blocks must accelerate together because the string is inextensible.

Another classic trap: assuming tension equals applied force. The string only “feels” the force needed to accelerate the rear block — never the full applied force unless the rear block is massless.

This template appeared in JEE Main 2024 (Shift 1, January 27), JEE Main 2023, and at least three NEET papers since 2020. Master this and we cover ~6-8 marks worth of NLM questions almost reflexively.

The rule of thumb: whenever blocks are connected by a string or in contact, find acceleration first using total mass, then isolate to find internal forces.

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