Connected BlocksJEE Main

Tension in a string towing a second block — interactive Physics simulation for IIT-JEE.

Concept

Treat the two blocks as one system to find the common acceleration, then cut the string and analyse a single block to find the tension. The tension only has to move the trailing block (and beat its friction), so it is always less than F.

Key formula

a = \frac{F - \mu(m_1+m_2)g}{m_1+m_2}, \qquad T = m_2a + \mu m_2 g = F\,\frac{m_2}{m_1+m_2} \;(\text{smooth})

Derivation

System: $F - \mu(m_1+m_2)g = (m_1+m_2)a$ .

Trailing block alone: $T - \mu m_2 g = m_2 a$ . Substituting a shows that on a smooth floor $T = F\frac{m_2}{m_1+m_2}$ — the string carries exactly the fraction of F needed for the mass behind it.

Scenarios to explore

Connected Blocks — Tension in a string towing a second block, with friction.

Real-world applications

Train couplings — tension drops toward the rear car by car.
Tug-of-war rope segments.
Towing chains and trailer hitch loads.

JEE exam tips

n identical blocks: tension after the k-th block from the front = F·(n−k)/n on smooth ground.
If F is applied to the LIGHTER block, tension is larger — check which side is pulled.
System first, then isolate — two quick equations beat four simultaneous ones.

Common mistakes

Setting T = F — the string does not transmit the full applied force.
Forgetting friction acts on BOTH blocks in the system equation.
Different accelerations for the two blocks (inextensible string ⇒ same a).

Exam traps to avoid

With friction, T ≠ F·m₂/(m₁+m₂) — the simple fraction only holds on smooth floors.
Below the friction threshold everything is static and tension can be ambiguous — exams stick to the moving case.