A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m s$$^{-1}$$, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is _______ J. (Given $$g = 10$$ m s$$^{-2}$$).

The tube is a vertical circle of radius $$R = 15 \text{ cm} = 0.15 \text{ m}$$. Take the bottom‐most point of the circle as the reference level of gravitational potential energy (GPE).

Initial (block released at the top)
Height of the top above the bottom is the diameter, $$h = 2R = 0.30 \text{ m}$$. Initial speed is $$v_0 = 22 \text{ m s}^{-1}$$.

Initial kinetic energy $$K_i = \tfrac12 m v_0^{2} = \tfrac12 (1) (22)^{2} = \tfrac12 (484) = 242 \text{ J}$$.

Initial gravitational potential energy $$U_i = mgh = (1)(10)(0.30) = 3 \text{ J}$$.

Hence the initial total mechanical energy $$E_i = K_i + U_i = 242 + 3 = 245 \text{ J}$$.

Final (block finally comes to rest at the bottom)
Final speed $$v_f = 0$$, so $$K_f = 0$$. At the bottom, $$U_f = 0$$ by our choice of reference. Therefore the final mechanical energy $$E_f = 0 + 0 = 0 \text{ J}$$.

Work done by the tube (friction)
For non-conservative forces, $$W_{\text{tube}} = E_f - E_i$$.

Substituting, $$W_{\text{tube}} = 0 - 245 = -245 \text{ J}$$.

The negative sign shows that the tube removes energy from the block. Thus the magnitude of the work done by the tube on the block is $$245 \text{ J}$$.

Answer: $$245 \text{ J}$$.