Practice question

Given: A 5 kg block on a 37° incline ( μ_k = 0.2 ) is pulled upward by a 7 kg mass over a pulley. What is the tension? (Take g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8 ) These values define the system as per NCERT data. Formula: For 7 kg : 7g - T = 7a Rightarrow 70 - T = 7a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5 kg : T - mg sin 37° - f_k = 5a . N = mg cos 37° = 5 × 10 × 0.8 = 40 N . Friction: f_k = 0.2 × 40 = 8 N . mg sin 37° = 50 × 0.6 = 30 N . Net force: T - 30 - 8 = 5a Rightarrow T - 38 = 5a . Solve: 70 - T = 7a, T - 38 = 5a . Substitute: 70 - (5a + 38) = 7a Rightarrow 70 - 38 - 5a = 7a Rightarrow 32 = 12a . a = 32/12 approx 2.67 m/s² . T - 38 = 5 × 2.67 Rightarrow T - 38 approx 13.35 Rightarrow T approx 51.35 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.