Practice question

Given: A 8.5 kg block on a horizontal surface ( μ_k = 0.25 ) is pulled by a 4.5 kg mass over a pulley. A 15 N force opposes the 8.5 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4.5 kg : 4.5g - T = 4.5a Rightarrow 45 - T = 4.5a. This is standard NCERT relation. Substitution & Calculation: For 8.5 kg : T - f_k - 15 = 8.5a . Normal: N = mg = 8.5 × 10 = 85 N . Friction: f_k = 0.25 × 85 = 21.25 N . Net force: T - 21.25 - 15 = 8.5a Rightarrow T - 36.25 = 8.5a . Solve: 45 - T = 4.5a, T - 36.25 = 8.5a . Substitute: 45 - (8.5a + 36.25) = 4.5a Rightarrow 45 - 36.25 - 8.5a = 4.5a Rightarrow 8.75 = 13a . a = 8.75/13 approx 0.67 m/s² . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.