A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon 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 density of air is constant.

Explanation

Let m, `V, rho_(He)` denote respectively the mass, volume and density of helium balloon and `rho_("air")` be density of air. Volume V of balloon displaces volume of air.

So, `V(rho_("air") - rho_("He")) g = m = a" "......(i)`
Integrating with respect to t,
`V(rho_("air") - rho_("He") gt = m v`
`implies 1/2 mv^2 = 1/2 m (V^2)/(m^2) (rho_("air") - rho_("He")) g^2 t^2 = 1/(2m) V^2 (rho_("air") - rho_("He"))^2 g^2 t^2" ".....(ii)`
If the balloon rises to a height h , from `s = ut + 1/2 at^2`,
We get `h = 1/2 (V(rho_("air") - rho_("He")))/(m) gt^2" ".....(iii)`
From Eqs. (iii) and (ii)
`1/2 mv^2 = [V (rho_(a) - rho_(He))g][1/(2m) V (rho_a - rho_(He))gt^2]" " = V (rho_a - rho_(He)) gh`
Rearranging the terms.
`implies 1/2 mv^2 + V rho_(He) gh = V rho_("air") hg implies KE_("baloon") + PE_("baloon")` = change in PE of air.
So, as the balloon goes up, an equal volume of air comes down, increase in PE and KE of the balloon is at the cost of PE of air [which come down]