= Solution
The component of <magnetic force> along the slope has magnitude $IBs\cos\phi$. The rate at which the sliding bar does work against this magnetic braking is
$$
P_{\mathrm{brake}}=IBs v\cos\phi=I\mathcal E.
$$
By <Ohm's law>, $\mathcal E=IR$, so
$$
\boxed{P_{\mathrm{brake}}=I^2R=\frac{B^2s^2v^2\cos^2\phi}{R}.}
$$
This exactly equals the <Joule heating> in the <resistor> in the negligible-<self-inductance> approximation. The <normal force> does no work because it is perpendicular to the <velocity>.
The gravitational <power> is instead $mgv\sin\phi$. By the <work-energy theorem>,
$$
mgv\sin\phi=I^2R+\frac{d}{dt}\left(\frac12mv^2\right).
$$
It equals the <Joule heating> only at constant speed. For a concrete counterexample, during the initial acceleration the <kinetic energy> increases, so some gravitational <power> remains mechanical.
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