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A ballon filled with helium rises agains...

A ballon filled with helium rises against gravity increasing its potential energy . The speed of thee ballon 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 .

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m = Mass of ballon
V = Volume of ballon
`rho_("He") ` = Density of helium
`rho_("air")` = Denity of air
Volume V of ballon displaces volume V of air .
So , `V(rho_(air)-rho_(He)) g = ma = m (dv)/(dt)" "…(i)`
Integrating with respect to t ,
` rArr V(rho_("air") -rho_("He")) "gt" = mv " "...(ii)`
`1/2 mv^(2) =1/2 m ((V^(2))/(m^(2))) (rho_(air)-rho_("He"))^(2) g^(2)t^(2)`
`= 1/(2m) V^(2)(rho_(air)-rho_(He))^(2) g^(2)t^(2)`
If the ballon rises to a height h ,
from s = ut + `1/2 at^(2)`
` :. h = 1/2 "at"^(2)`
`=1/2(V(rho_(air)-rho_(He)))/m "gt"^(2) ( :. u = 0 ) ....(iii)`
From (iii) and (ii)
`1/2 mv^(2) = [V (rho_(a)-rho_(He))"gt"^(2)][1/(2m)V(rho_(air)-rho_(He))"gt"^(2)]`
` =V(rho_(a)-rho_(He))gh`
Therefore ,
`1/2 mv^(2) +V_(rho_(He))=gh =V_("pair") hg `
`KE_("ballon")+PE_("ballon") ` = Change in PE of air .
So, as the ballon goes up , an equal volume of air comes down , increase in PE and kE of the ballon is equal to decrease inPE of air .
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