Practice question

Given: A 5 kg block on a horizontal surface ( μ_k = 0.2 ) is pulled by a 8 kg mass over a pulley. A 10 N force opposes the 5 kg block. What is the tension? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 8 kg : 8g - T = 8a Rightarrow 80 - T = 8a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5 kg : T - f_k - 10 = 5a . Normal: N = mg = 5 × 10 = 50 N . Friction: f_k = 0.2 × 50 = 10 N . Net force: T - 10 - 10 = 5a Rightarrow T - 20 = 5a . Solve: 80 - T = 8a, T - 20 = 5a . Substitute: 80 - (5a + 20) = 8a Rightarrow 80 - 20 - 5a = 8a Rightarrow 60 = 13a . a approx 4.62 m/s², T - 20 = 5 × 4.62 Rightarrow T - 20 approx 23.1 Rightarrow T approx 43.1 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.