The heat developed in the wire is used to melt the ice. Let the wire move with a constant velocity $v$ relative to the ice. In time $dt$, the wire travels a distance $dx = v dt$ and melts a volume of ice equal to the volume of the wire section passing through it.

Geometry:
The effective length of the wire interacting with the ice is the thickness $h$. The diameter of the wire is $2r$.
Volume melted in time $dt$: $$ dV = (2r) \cdot h \cdot dx $$ Mass melted: $$ dm = \text{density} \times dV = d \cdot 2rh \cdot dx $$ Heat required for melting (where $L$ is specific latent heat): $$ dQ_{req} = dm \cdot L = 2 d L h r dx $$

Electrical Power:
The resistance of the wire section of length $h$ is given by $R = \rho \frac{h}{\pi r^2}$.
Heat generated by the current in time $dt$: $$ dQ_{gen} = \frac{V^2}{R} dt = \frac{V^2 \pi r^2}{\rho h} dt $$

Energy Balance:
Equating the heat generated to the heat required: $$ \frac{V^2 \pi r^2}{\rho h} dt = 2 d L h r dx $$ Solving for the velocity of the wire $v = \frac{dx}{dt}$: $$ v = \frac{V^2 \pi r^2}{2 d L h^2 r \rho} = \frac{\pi r V^2}{2 d L h^2 \rho} $$ This is the maximum speed at which the wire can melt through the ice.

To cut a rectangular log of width $b$ (transverse cut), the wire must traverse the distance $b$. The minimum time $t$ is: $$ t = \frac{\text{Distance}}{v} = \frac{b}{v} $$ $$ t = \frac{b}{\left( \frac{\pi r V^2}{2 d L h^2 \rho} \right)} = \frac{2 L \rho d b h^2}{\pi r V^2} $$

Now the log moves with velocity $\vec{v}_{log} = v_x \hat{i}$ (along x-axis). The wire moves with velocity $\vec{v}_{wire} = v_y \hat{j}$ (along y-axis). The cutting process depends on the relative velocity of the wire with respect to the ice: $$ \vec{v}_{rel} = \vec{v}_{wire} – \vec{v}_{log} = v_y \hat{j} – v_x \hat{i} $$ The magnitude of this relative velocity represents the speed at which the wire cuts through the ice material: $$ |\vec{v}_{rel}| = \sqrt{v_x^2 + v_y^2} $$

Efficiency Condition:
“Most energy efficient cutting” implies minimizing the heat wasted, which corresponds to completing the cut as fast as the physics of melting allows. This occurs when the relative speed is at its maximum possible value derived in Part (a), denoted as $v_{max}$: $$ v_{max} = \frac{\pi r V^2}{2 d L h^2 \rho} $$ Therefore, we set the relative speed equal to this limit: $$ \sqrt{v_x^2 + v_y^2} = v_{max} $$ Squaring both sides: $$ v_x^2 + v_y^2 = \left( \frac{\pi r V^2}{2 d L h^2 \rho} \right)^2 $$ Solving for the required wire velocity $v_y$: $$ v_y = \sqrt{ \left( \frac{\pi r V^2}{2 d L h^2 \rho} \right)^2 – v_x^2 } $$