$A$ balloon filled with helium rises against gravity,increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that the density of air is constant.

(N/A) The mechanical energy of the balloon alone is not conserved because an external buoyant force acts on it. The work done by the buoyant force is responsible for the increase in the total mechanical energy (kinetic energy + potential energy) of the balloon.
Let $m$ be the mass of the balloon,$V$ be its volume,$\rho_{He}$ be the density of helium,and $\rho_{air}$ be the density of air.
The net upward force on the balloon is $F_{net} = V(\rho_{air} - \rho_{He})g - mg = ma$.
As the balloon rises to a height $h$,the work done by the buoyant force is $W_b = V \rho_{air} g h$.
The change in mechanical energy is $\Delta E = \Delta K + \Delta U = \frac{1}{2}mv^2 + mgh$.
Using the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W_{net} = [V(\rho_{air} - \rho_{He})g - mg]h = \frac{1}{2}mv^2$.
Rearranging this,we get $V \rho_{air} g h = \frac{1}{2}mv^2 + mgh$.
This shows that the work done by the buoyant force $(V \rho_{air} g h)$ is exactly equal to the increase in the total mechanical energy of the balloon.