In a vertical plane, a closed circular tube of radius 10 cm...

Let the lowest point of the tube be the reference level for gravitational potential energy ($h = 0$).

At the highest point, the height of the block is $h = 2R = 2 \times 0.1 = 0.2 \text{ m}$.

The initial mechanical energy of the block is the sum of its kinetic and potential energies:

$E_i = K_i + U_i = \dfrac{1}{2}mv^2 + mgh$

Substituting the given values ($m = 0.1 \text{ kg}$, $v = 5.0 \text{ m/s}$, $g = 10 \text{ m/s}^2$):

$E_i = \dfrac{1}{2}(0.1)(5.0)^2 + (0.1)(10)(0.2)$

$E_i = 1.25 + 0.20 = 1.45 \text{ J}$

The block finally comes to rest at the lowest point, so its final velocity is zero and its final height is zero.

The final mechanical energy is:

$E_f = K_f + U_f = 0 + 0 = 0 \text{ J}$

The work done by the tube on the block is equal to the work done by friction (since the normal force is perpendicular to the displacement and does zero work). By the work-energy theorem, the work done by non-conservative forces is equal to the change in the total mechanical energy of the block:

$W = \Delta E = E_f - E_i$

$W = 0 - 1.45 = -1.45 \text{ J}$