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

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

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)

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$\huge{AηsωeR} ✍$

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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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