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Work, Energy, and Power — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
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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Answer

As the dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vpg = Volume of air displaced × net density upward × g$=\text{V}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{g}(\text{upward})$
Let a be the upward acceleration on balloon then,$\text{ma}=\text{V}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{g}\ ...(\text{i})$
V = Volume of air displacement by balloon = Volume of balloon $p_{air}$ = density of air $p_{He}$ = density of helium$\text{m}\frac{\text{dv}}{\text{dt}}=\text{V}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{g}$
$\text{m dv}=\text{V}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{g}\cdot\text{dt}$
Integrating both sides $\text{mv}=\text{V}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{gt}$$\text{v}=\frac{\text{V}}{\text{m}}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{gt}$
KE of balloon $=\frac{1}{2}\text{mv}^2$$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{m}\frac{\text{v}^2}{\text{m}^2}(\text{p}_{\text{air}}-\text{p}_\text{He})^2\text{g}^2\text{t}^2$
$=\frac{\text{V}^2}{2\text{m}}(\text{p}_{\text{air}}-\text{p}_\text{He})^2\text{g}^2\text{t}^2\ ....(\text{ii})$
If the balloon rises to a height h, from (i)$\text{a}=\frac{\text{V}}{\text{m}}(\text{p}_{\text{air}}-\text{p}_\text{He})\text{g}$
$\text{h}=\text{ut}+\frac{1}{2}\text{at}^2=0\cdot\text{t}+\frac{1}{2}\Big[\frac{\text{V}}{\text{m}}(\text{p}_\text{air}-\text{p}_\text{He})\Big]\text{t}^2$
$\therefore\ \text{h}=\frac{\text{V}}{2\text{m}}(\text{p}_\text{air}-\text{p}_\text{he})\text{gt}^2\ ...(\text{iii})$
From (ii) and (iii) rearranging the terms of (ii) according to h in (iii)$\frac{1}{2}\text{mv}^2=\Big\{\frac{\text{V}}{2\text{m}}(\text{p}_\text{air}-\text{p}_\text{He})\text{gt}^2\Big\}\cdot\text{V}(\text{p}_\text{air}-\text{p}_{\text{He}})\text{g}$
$\frac{1}{2}\text{mv}^2=\{\text{h\}}\cdot\text{V}(\text{p}_\text{air}-\text{p}_{\text{He}})\text{g}$
$\frac{1}{2}\text{mv}^2=\text{V}\cdot(\text{p}_\text{air}-\text{p}_{\text{He}})\text{gh}$
$\frac{1}{2}\text{mv}^2=\text{V}\text{p}_\text{air}\text{gh}-\text{V}\text{p}_{\text{He}}\text{gh}$
$\frac{1}{2}\text{mv}^2+\text{p}_{\text{He}}\text{Vgh}=\text{p}_\text{air}\text{Vgh}$
$\text{KE}_{\text{ballon}}+\text{PE}_{\text{ballon}}=\text{Changein PE of air}$
So, as the balloon goes up, an equal volume of air comes down, increases in PE and KE of the balloon is at cost of PE of air (which comes down).

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