Practice question

Given: A 5 kg mass on a horizontal surface ( μ_k = 0.4 ) is pulled by a 3 kg mass over a pulley with a 10 N force aiding the 5 kg mass. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 3 kg : 3g - T = 3a Rightarrow 30 - T = 3a. This is standard NCERT relation. Substitution & Calculation: For 5 kg : T + 10 - f_k = 5a, f_k = 0.4 × 5 × 10 = 20 N . T + 10 - 20 = 5a Rightarrow T - 10 = 5a . Solve: 30 - T = 3a, T - 10 = 5a Rightarrow 30 - (5a + 10) = 3a . 30 - 10 - 5a = 3a Rightarrow 20 = 8a Rightarrow a = 2.5 m/s² . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.