1. Interpret the motion

A block is introduced at the top of a rough vertical circular tube with speed $22\,\text{m/s}$.

Because the tube is rough, friction dissipates mechanical energy. After some to-and-fro motion, it finally comes to rest in the bottom region of the tube after 5 oscillations.

We are asked for the work done by the tube on the block.

The tube exerts:

• normal reaction
• friction

The normal reaction does no net work (it is perpendicular to instantaneous displacement along the tube), so the total work done by the tube equals the work done by friction, i.e. the change in mechanical energy due to non-conservative force.

2. Use work-energy relation

Work done by the tube on the block is $W_{\text{tube}} = (K_f + U_f) - (K_i + U_i).$

Take gravitational potential energy zero at the bottom of the tube.

Then:

• At the top, height above bottom = $2R = 2(0.15)=0.30\,\text{m}$
• At the bottom, height = $0$

So initial and final energies are:

Initial energy

$K_i = \frac12 mv^2 = \frac12(1)(22)^2 = 242\,\text{J}$ $U_i = mg(2R) = (1)(10)(0.30) = 3\,\text{J}$

Thus, $E_i = K_i + U_i = 242 + 3 = 245\,\text{J}.$

Final energy

The block stops at the bottom region, so $K_f = 0, \qquad U_f = 0.$

Hence, $E_f = 0.$

Therefore, $W_{\text{tube}} = E_f - E_i = 0 - 245 = -245\,\text{J}.$

3. Meaning of the sign

The tube removes energy from the block, so the work done by the tube on the block is negative.

Thus the actual work done is $\boxed{-245\,\text{J}}.$

If the question expects the magnitude of energy dissipated by the tube, that would be $245\,\text{J}$.

4. Comparison with stored answer

Stored correct answer is $245$.

But strictly, the work done by the tube on the block is negative: $\boxed{-245\,\text{J}}.$ The value $245\,\text{J}$ is only the magnitude of energy lost.