Two blocks m₁ and m₂ on smooth horizontal surface, connected by a light inextensible string. Force F applied on m₂ (which is at the far end from m₁). Since string is inextensible: both blocks have the same acceleration a. System: F = (m₁+m₂)a → a = F/(m₁+m₂). Tension T: isolate m₁ (free block, only tension pulls it). T = m₁a = m₁F/(m₁+m₂). Alternatively isolate m₂: F − T = m₂a → T = F − m₂a = F − m₂F/(m₁+m₂) = m₁F/(m₁+m₂) ✓. Note: if F is applied on m₁, the string now PUSHES from behind — wait, a string cannot push. If you apply F on m₁ and m₂ is connected by string: m₁ pushes the string (compression?) but a string cannot take compression. In this case, there is NO tension — it's a contact force scenario, not a string tension scenario. String only if pulling.

• Comparison: contact force vs. tension for two-block systems. If blocks are just in contact (no string) and F applied on m₁ pushing m₂: contact force R = m₂F/(m₁+m₂). If blocks are connected by string and F pulls m₂ (dragging m₁ behind): T = m₁F/(m₁+m₂). The formulas swap m₁ and m₂ in the numerator! Contact force = m_ahead; tension = m_behind. This is the most NEET-relevant distinction between these two scenarios.
• Tension is the same throughout a massless string in a simple horizontal system. If the string has TWO segments (e.g., m₁ — string1 — pulley — string2 — m₂), and pulley is frictionless: T₁ = T₂ (same tension in both segments). The string changes direction at the pulley but the tension magnitude is preserved. If pulley has friction: T₁ ≠ T₂ — this is NOT typically tested in NEET at this level.
• System check: if the string were to go slack, tension T = 0. This occurs when m₁ = 0 (nothing to pull), or when the applied force F is not in the direction that would create tension. If the acceleration of the blocks becomes such that m₁ would accelerate FASTER than m₂ (e.g., m₁ is on an incline and slides down faster than the applied force drives the system), the string goes slack — analyze the blocks independently in this case.

Three blocks m₁, m₂, m₃ connected as: m₁ — T₁ — m₂ — T₂ — m₃ — F. Force F applied on m₃ (far right), string T₁ is the left string, string T₂ is the middle string. All blocks accelerate at a = F/(m₁+m₂+m₃). T₁ (string between m₁ and m₂): isolate m₁ alone. Only T₁ acts on m₁. T₁ = m₁a = m₁F/(m₁+m₂+m₃). T₂ (string between m₂ and m₃): isolate m₁ and m₂ as a sub-system. T₂ pulls them forward. T₂ = (m₁+m₂)a = (m₁+m₂)F/(m₁+m₂+m₃). Since m₁+m₂ > m₁: T₂ > T₁. The string nearest F carries the highest tension.

• Alternative formula for T in any string in a multi-block chain: T_n = (sum of masses of all blocks that the string n must pull) × a. The string at position n must pull all blocks 'behind' it (farther from F). String T₁ pulls only m₁. String T₂ pulls m₁ and m₂. Hence T₂ > T₁. For a chain of n blocks, the leftmost string (farthest from F) has the lowest tension; the rightmost string (nearest F) has the highest tension.
• NEET trap scenario: 'same magnitude force F applied on m₁ (opposite end) — find tensions.' a = F/(m₁+m₂+m₃) (same). Now: string T₁ (between m₁ and m₂): F is on m₁, so T₁ must pull m₂ and m₃ forward. T₁ = (m₂+m₃)a — highest tension near the new applied force end. T₂ (between m₂ and m₃): pulls only m₃. T₂ = m₃a — lowest tension. Everything flips compared to the previous case. The string closest to the applied force always has the highest tension regardless of which end F is applied.
• Numerical memory aid: three equal masses m each, total force F: a = F/3m. T₁ (string farthest from F) = m × F/3m = F/3. T₂ (string nearest to F) = 2m × F/3m = 2F/3. Ratio T₂:T₁ = 2:1. For n equal masses: strings have tensions F/n, 2F/n, 3F/n, ... (n-1)F/n, (decreasing from F as you go away from applied force). The string ratio forms an arithmetic progression for equal masses.

Two blocks m₁ and m₂ connected by string on rough surface (μ_k). Force F on m₂. System: F − μ_k m₁g − μ_k m₂g = (m₁+m₂)a → a = [F − μ_k(m₁+m₂)g]/(m₁+m₂). Tension T₁: isolate m₁. Forces: T₁ (forward) and friction f₁ = μ_k m₁g (backward). T₁ − μ_k m₁g = m₁a → T₁ = m₁(a + μ_k g). This equals m₁ × F/(m₁+m₂) — same as the smooth case formula. The friction on the isolated block must be explicitly added; it does not vanish just because system acceleration already accounts for it.

• Why T = m₁ × F/(m₁+m₂) independent of friction (as long as μ is same for both): substitute a = [F − μ(m₁+m₂)g]/(m₁+m₂) into T = m₁(a + μg): T = m₁ × {[F − μ(m₁+m₂)g]/(m₁+m₂) + μg} = m₁ × {F/(m₁+m₂) − μg + μg} = m₁F/(m₁+m₂). The friction terms cancel! This is only true when BOTH blocks have the same coefficient of friction. If μ₁ ≠ μ₂, the cancellation does not occur and T must be computed from T = m₁(a + μ₁g) using a = [F − μ₁m₁g − μ₂m₂g]/(m₁+m₂).
• Different friction coefficients: F on m₂, m₁ and m₂ connected by string. Friction on m₁ = μ₁m₁g, on m₂ = μ₂m₂g. System: a = [F − μ₁m₁g − μ₂m₂g]/(m₁+m₂). Tension T: isolate m₁: T = m₁a + μ₁m₁g = m₁(a + μ₁g). Substituting: T = m₁ × [(F − μ₁m₁g − μ₂m₂g)/(m₁+m₂) + μ₁g] = m₁[F + μ₁m₂g − μ₂m₂g]/(m₁+m₂) = m₁[F + (μ₁−μ₂)m₂g]/(m₁+m₂). Now T ≠ m₁F/(m₁+m₂) when μ₁ ≠ μ₂.
• String tension on incline (m₁ on incline angle θ, connected to hanging m₂ via string over pulley): different from pure horizontal. This is the Atwood machine with one mass on incline. Force along incline: m₁g sinθ (down slope) opposes tension T if m₂ is causing upward motion of m₁. System: m₂g − m₁g sinθ = (m₁+m₂)a (assuming m₂ descends). T = m₁(a + g sinθ). Rough incline adds μm₁g cosθ term.