A 4 kg block rests on a rough horizontal surface with a coefficient of kinetic friction of 0.5 . It is connected by a light inextensible string passing over a smooth pulley to a hanging 6 kg block. If the system is released from rest, what is the distance descended by the hanging block in the first 2 s ? (Take g = 10 m s ⁻² )

Step-by-step solution

Let the acceleration of the system be a and the tension in the string be T . For the hanging block ( 6 kg ), the driving force is its weight: m₂ g = 6 10 = 60 N The equation of motion is: 60 - T = 6a For the block on the horizontal surface ( 4 kg ), the opposing kinetic friction is: f_k = _k m₁ g = 0.5 4 10 = 20 N The equation of motion is: T - 20 = 4a Adding the two equations to find the common acceleration: 60 - 20 = (6 + 4)a 40 = 10a a = 4 m s ⁻² Now, using the kinematic equation for uniformly accelerated motion