Practice question

Given: A 4.6 kg block on a horizontal surface ( μ_k = 0.2 ) is connected to a 6.6 kg mass over a pulley. A 9 N force aids the 4.6 kg block. What is the acceleration? (Take g = 10 m/s² ) Formula: For 6.6 kg : 6.6g - T = 6.6a Rightarrow 66 - T = 6.6a. Substitution & Calculation: The 6.6 kg mass descends, pulling the 4.6 kg block with an aiding force. . For 4.6 kg : T + 9 - f_k = 4.6a . Normal: N = mg = 4.6 × 10 = 46 N . Friction: f_k = 0.2 × 46 = 9.2 N . Net force: T + 9 - 9.2 = 4.6a Rightarrow T - 0.2 = 4.6a . Solve: 66 - T = 6.6a, T - 0.2 = 4.6a . Substitute: 66 - (4.6a + 0.2) = 6.6a Rightarrow 66 - 0.2 - 4.6a = 6.6a Rightarrow 65.8 = 11.2a . a = 65.8/11.2 approx 5.88 m/s² . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.