A block of mass 8 kg is placed on a smooth horizontal floor. It is pulled by a uniform heavy rope of mass 4 kg and length 2 m attached to it. A constant horizontal force of 60 N is applied to the free end of the rope. The tension in the rope at a distance of 0.5 m from the block is:

Step-by-step solution

The total mass of the system is the sum of the mass of the block and the mass of the rope. M_ total = 8 kg + 4 kg = 12 kg The common acceleration of the system is: a = F M_ total = 60 12 = 5 m s ⁻² We need to find the tension at a point in the rope which is 0.5 m away from the block. The mass of this 0.5 m segment of the rope is: m_ segment = ( 4 kg 2 m ) 0.5 m = 1 kg The total mass being pulled by the tension at this point (the trailing mass) includes the block and the 0.5 m segment of the rope: M_ trailing = 8 kg