Question

①• A force is applied on a body of mass 20 kg moving with a velocity of 40 m/s. The body attains a velocity of
50 m/s in 2 s. Calculate the work done on the body.

②• A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy? (Take g = 10 m/s²)

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Answer 1

ANSWER -2

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

★ Formula Used :

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

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

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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

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

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

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

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

This is because, Kg m²/sec² = Joules

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

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

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

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

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

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

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

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

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

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

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

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

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

Since, J / sec = Watt.

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

★ More Formulas to Know :

$\begin{gathered}\\\;\bf{\leadsto\;\;\;P\;=\;Fv}\end{gathered}$

$\begin{gathered}\\\;\bf{\leadsto\;\;\;\omega\;=\;\omega_{0}\;+\;\alpha t}\end{gathered}$

$\begin{gathered}\\\;\bf{\leadsto\;\;\;\theta\;=\;\omega_{0} t\;+\;\dfrac{1}{2}\:\alpha\:t^{2}}\end{gathered}$

$\begin{gathered}\\\;\bf{\leadsto\;\;\;\omega^{2}\;-\;\omega_{0} ^{2}\;=\;2 \alpha \theta}\end{gathered}$

$\begin{gathered}\\\;\bf{\leadsto\;\;\;E\;=\;mc^{2}}\end{gathered}$

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-1  }}$

A force is applied on a body of mass 20 kg moving with a velocity of 40 m/s. The body attains a velocity of

50 m/s in 2 s. Calculate the work done on the body.

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$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{initial \: velocity, \: u \: = 40 \: m \: per \: sec.} \\ &\sf{mass, \: m = 20 \: kg}\\ &\sf{time \: taken, \: t \: = 2 \: sec.}\\ &\sf{final \: velocity, \: v \: = 50 \: m \: per \: sec.} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{work \: done} \\ \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Formula \: used:- \begin{cases} &\sf{v = u + at} \\ &\sf{Force = m × a}\\ &\sf{ {v}^{2} - {u}^{2} = 2as}\\ &\sf{Work \: done = Force × Displacement} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\large\underline\red{\bold{❥︎Step :- 1 }}$

❥︎We know,

$\sf \:  ⟼v = u + at$

$\sf \:  ⟼v - u = at$

$\bf\implies \:a \: = \dfrac{v - u}{t} \sf \:  ⟼(1)$

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$\large\underline\red{\bold{❥︎Step :- 2 }}$

$\sf \:  ⟼Force = m \: × \: a$

On substituting the value from equation (1), we get

$\sf \:  ⟼Force = m \: × \: \dfrac{v - u}{t} \sf \:  ⟼(2)$

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

$\sf \:  ⟼ {v}^{2} - {u}^{2} = 2as$

$\sf \:  ⟼s = \dfrac{ {v}^{2} - {u}^{2} }{2a}$

On substituting the value of 'a' from equation (1), we get

$\sf \:  ⟼s = \dfrac{ {v}^{2} - {u}^{2} }{2 \times \dfrac{v - u}{t} }$

$\sf \:  ⟼s = \dfrac{( \cancel{v - u)}(v + u)}{2 \times (\cancel{v - u)}} \times t$

$\sf \:  ⟼s = \dfrac{t}{2} \times (v - u)\sf \:  ⟼(3)$

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$\large\underline\red{\bold{❥︎Step :- 4 }}$

$\sf \:  ⟼Work \: done \: = \: Force \: × \: Displacement$

$\sf \:  ⟼Work \: done \: = m \times \dfrac{v - u}{t} \times \dfrac{t \times (v + u)}{2}$

$\sf \:  ⟼Work \: done \: = \dfrac{m}{2} ( {v}^{2} - {u}^{2} \: )$

$\sf \:  ⟼Work \: done \: = \dfrac{20}{2} ( {(50)}^{2} - {(40)}^{2} \: )$

$\sf \:  ⟼Work \: done \: = 10 \times 900 = 9000$

$\bf\implies \:  ⟼Work \: done \: = 9000 \: joules.$

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$\large\underline\blue{\bold{Given \:  Question :-2  }}$

A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy? (Take g = 10 m/s²)

─━─━─━─━─━─━─━─━─━─━─━─━─

$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf given \: - \begin{cases} &\sf{mass \: of \: girl, \: m = 20 \: kg} \\ &\sf{height \: of \: first \: floor \: from \: ground \: = 4 \: m}\\ &\sf{height \: of \: third \: floor \: from \: ground \: = 12 \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{incrase \: in \: gravitational \: potential \: energy} \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\sf \: Gravitational \: potential \: energy \: at \: first \: floor = mgh_1$

$\sf \:  = 35 \times 10 \times 4$

$\sf \:  = 1400 \: joules$

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$\sf \:Gravitational \: potential \: energy \: at \: {3}^{rd} \: floor = mgh_2$

$\sf \:  = 35 \times 10 \times 12 = 4200 \: joules$

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$\sf \:So, \: change \: in gravitational \: potential \: energy$

$\bf \:  ⟼ 4200 - 140 = 2800 \: joules.$

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