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प्रश्न
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.
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उत्तर
Because the pulling viscous force of the air on the balloon is ignored, there is
Net Buoyant Force = `Vpg`
The volume of air displaced × net density upward × g
= `V(p_(ar) - p_(He)) g(upward)`
Let a be the upward acceleration on the balloon then
`ma = V(p_(ag) - p_(He)) g`
Where m = mass of the balloon
V = Volume of air displacement by balloon
= Volume of balloon
`p_(air)` = density of air
`p_(He)` = density of helium
`m (dv)/(dt) = V * (p_(ar) - p_(He)) g`
`mdv = V * (p_(air) - p_(He)) g * dt`
Integrating both sides `mv = V * (p_(ar) - p_(He)) g t`
`v = V/m (p_(air) - p_(He)) g t`
KE of balloon = `1/2 mv^2`
`1/2 mv^2 = 1/2 m v^2/m^2 (p_(air) - p_(He))^2g^2t^2`
= `V^2/(2m) (p_(air) - p_(He))^2g^2t^2` ......(ii)
If the balloon rises to a height h, from (i)
`a = V/m (p_(ai) - p_(he)) g`
`h = ut + 1/2 at^2 = 0.t + 1/2 [V/m (p_(ai) - p_(Bb))g]t^2`
∴ `h = V/(2m) (p_(ar) - p_(He)) g t^2`
Rearranging the words of (ii) according to h in (ii) and (iii) (iii)
`1/2 mv^2 = {V/(2m) (P_(air) - p_(He)) g t^2} * V(P_(air) - p_(He)) g`
`1/2 mv^2 = {h} * V (p_(air) - p_(He)) g`
`1/2 mv^2 = V * (p_(at) - p_(He)) gh`
`1/2 mv^2 = Vp_(air) gh - Vp_(He) = gh`
`1/2 mv^2 + p_(He)V gh = p_(av) V gh`
`KE_(balloon) + PE_(ballon)` = Change in PE of air.
As a result, as the balloon rises, an equivalent volume of air falls, increasing the balloon's PE and KE at the expense of the air's PE (which come down).
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