JEE Main Physics question
A block of mass 5 kg, resting on a horizontal surface, is connected by a cord passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the first block and the surface is 0.5. Find the tension in the cord. (g = 9.8 ms⁻²)
Answer: (C)
Solution
a = (m₂g−μm₁g)/(m₁+m₂) = (5g−2.5g)/10 = 0.25g = 2.45 ms⁻². T = m₂(g−a) = 5(9.8−2.45) = 36.75 N.