A small block is released from rest from a height of 20 m on a frictionless curved track. At the lowest point, it passes through a short rough horizontal patch where it loses 40 % of its kinetic energy due to friction. It then continues to move up another frictionless curved track. The maximum vertical height reached by the block on the second track is:

Step-by-step solution

By the principle of conservation of mechanical energy, the kinetic energy of the block at the lowest point before entering the rough patch is equal to its initial potential energy. K₁ = mgh₁ The block loses 40 % of its kinetic energy on the rough horizontal patch. The remaining kinetic energy is: K₂ = K₁ - 0.40 K₁ = 0.60 K₁ K₂ = 0.60 mgh₁ As the block moves up the second frictionless track, its kinetic energy is converted into potential energy. At the maximum height h₂ , the potential energy is equal to K₂ . mgh₂ =