Practice question

Given: A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4 kg : 4g - T = 4a Rightarrow 40 - T = 4a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 7 kg : T - f_k - 14 = 7a . Normal: N = mg = 7 × 10 = 70 N . Friction: f_k = 0.3 × 70 = 21 N . Net force: T - 21 - 14 = 7a Rightarrow T - 35 = 7a . Solve: 40 - T = 4a, T - 35 = 7a . Substitute: 40 - (7a + 35) = 4a Rightarrow 40 - 35 - 7a = 4a Rightarrow 5 = 11a . a = 5/11 approx 0.45 m/s² . 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.