Question

Era, having mass 50 kg, climbs up from the first floor of her school at a height of 5.75 m above the
ground to the third floor at a height of 14.25 m above the ground in 19 s. Calculate -
a. Increase in her potential energy
b. Power consumed in doing so (g = 9.8 m/s?]
​

Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Question's\;\;Analysis\;\;:-}}}$

Here the concept of Potential Energy and Power has been used. We know that Energy is equal to Work done. So the increase in Potential Energy will be the work done by Era. Also we are given the time taken, so we can easily find both answers.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Potential\;Energy\;=\;\bf{mg\;\times\;height}}}$

$\\\;\boxed{\sf{Power,\;P\;\;=\;\bf{\dfrac{Work}{Time}}}}$

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★ Solution :-

Given,

» Mass which Era has = m = 50 Kg

» Height of the First Floor = h = 5.75 m

» Height of the Second Floor = h' = 14.25 m

» Time taken by Era for this = t = 19 sec

» Acceleration due to gravity = g = 9.8 m/sec²

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~ For the Initial Potential Energy of Era at the height of First Floor ::

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\bf{mg\;\times\;height}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\bf{mgh}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\bf{50\;\times\;9.8\;\times\;5.75}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\underline{\underline{\bf{2817.5\;\;Joules}}}}$

This is because, Kg m²/sec² = Joules

$\\\;\underline{\boxed{\tt{Initial\;\;Potential\;\;Energy\;\;of\;\;the\;\;Body\;=\;\bf{2817.5\;\;Joules}}}}$

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~ For the Final Potential Energy of Era at the height of Third Floor ::

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\bf{mg\;\times\;height}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\bf{mgh'}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\bf{50\;\times\;9.8\;\times\;14.25}}$

$\\\;\;\;\;\sf{:\rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\underline{\underline{\bf{6982.5\;\;Joules}}}}$

$\\\;\underline{\boxed{\tt{Final\;\;Potential\;\;Energy\;\;of\;\;the\;\;Body\;=\;\bf{6982.5\;\;Joules}}}}$

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a.) For the Increase in Potential Energy of Era :-

$\\\;\sf{:\Longrightarrow\;\;\;Increase\;in\;Potential\;Energy\;=\;\bf{Final\;P.E.\;-\;Initial\;P.E.}}$

$\\\;\sf{:\Longrightarrow\;\;\;Increase\;in\;Potential\;Energy\;=\;\bf{6982.5\;-\;2817.5}}$

$\\\;\bf{:\Longrightarrow\;\;\;Increase\;in\;Potential\;Energy\;=\;\bf{4165\;\;Joules}}$

$\\\;\large{\underline{\underline{\rm{Change\;in\;P.E.\;of\;Era\;is\;\;\boxed{\bf{4165\;\;Joules}}}}}}$

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b) For the Power consumed by Era :-

Since, Energy = Work

So, Potential Energy used by Era = Work

Then,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Power,\;P\;\;=\;\bf{\dfrac{Work}{Time}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Power,\;P\;\;=\;\bf{\dfrac{4165}{19}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Power,\;P\;\;=\;\bf{219.21\;\;Watt}}$

Since, J / sec = Watt.

$\\\;\large{\underline{\underline{\rm{Thus,\;power\;consumed\;by\;Era\;is\;\;\boxed{\bf{219.21\;\;Watt}}}}}}$

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★ More Formulas to Know :-

$\\\;\sf{\leadsto\;\;\;P\;=\;Fv}$

$\\\;\sf{\leadsto\;\;\;\omega\;=\;\omega_{0}\;+\;\alpha t}$

$\\\;\sf{\leadsto\;\;\;\theta\;=\;\omega_{0} t\;+\;\dfrac{1}{2}\:\alpha\:t^{2}}$

$\\\;\sf{\leadsto\;\;\;\omega^{2}\;-\;\omega_{0} ^{2}\;=\;2 \alpha \theta}$

$\\\;\sf{\leadsto\;\;\;E\;=\;mc^{2}}$