Question

An object of mass 40 kg is raised to a height of 5 m above the ground . what is it's potential energy ? If the object is allowed to fall, find it's kinetic energy when it is half-way down .
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Answer 1

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An object of mass 40 kg is raised to a height of 5 m above the ground . what is it's potential energy ? If the object is allowed to fall, find it's kinetic energy when it is half-way down

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$\large{\sf{Given,}}$

• Mass of the Object,M = 40 Kg

• Height,h = 5 m

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Work Done under the influence of gravity is known as Potential Energy

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To finD :

Potential Energy and Kinetic Energy

Mathematically, Potential Energy is given by :

$\boxed{\boxed{\tt{PE = Mgh}}}$

$\sf{Given}\begin{cases}\sf{M \colon Mass }\\\sf{g \colon Acceleration \ due \ to \ gravity}\\ \sf{h \colon height} \end{cases}$

Substituting the values,we get :

$\sf{\longrightarrow PE = 40 \times 10 \times 5} \\ \\ \sf{\longrightarrow PE = 40 \times 50} \\ \\ \longrightarrow \underline{\boxed{\sf{PE = 2000 \ J}}}$

The Total Work Done is 2000 J

When the objects is half way down,the height would be 2.5 m

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Note

According to Law of Conservation of Mechanical Energy,

Total PE = KE in the first half + PE in the second half

PE is not consider in first half :

The object is still on the ground frame. Thus,

h = 0m » PE = 0 J

KE is not considered in second half :

Generally during free fall,the final velocity of the object tends to zero. Thus,the Kinetic Energy at top is zero

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Now,

• PE in the first half would be the half of total potential energy i.e., it would be 1000 J

The same can be arrived mathematically by substituting M = 40 Kg,g = 10 m/s² and h = 2.5 m

Therefore,

$\sf{\longrightarrow \ 2000 = 1000 + KE} \\ \\ \sf{\longrightarrow KE = 2000 - 1000} \\ \\ \longrightarrow \ \underline{\boxed{\sf{KE = 1000 J}}}$

The Kinetic Energy of the object would be 1000 J

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Answer 2

❏ Question:-

An object of mass 40 kg is raised to a height of 5 m above the ground . what is it's potential energy ? If the object is allowed to fall, find it's kinetic energy when it is half-way down .

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❏ Solution:-

✦ Formula used:-

let, an object of mass m is raised to the height of h ,

➝ So, at the height of h ,

$\sf \longrightarrow\boxed{ P.E.= mgh}$

$\sf \longrightarrow \boxed{K.E.= 0}\:$

$\text{(as,velocity at the top most position becomes 0)}$

➝ Now, if the object is released to fall and if it fell to the height of d from the top most , then

$\sf \longrightarrow\boxed{ P.E.= mg(h-d)}$

$\sf \longrightarrow \boxed{K.E.= mgd}\:$

✦Fig :-

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✦Given :-

• mass(m) = 40 kg .

(m) = 40 kg .• height(h) = 5m .

(h) = 5m .• g = 9.8 m/s²

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➝The P.E.at point A:-

$\sf \implies P.E._{A}=mgh$

$\sf \implies P.E._{A}=40\times9.8\times5$

$\sf \implies\boxed{\large{\red{ P.E._{A}\:=\:1960 \: Joule}}}$

➝The K.E.at point A:-

$\sf \implies\underline{ K.E._{A}=0}$

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Now, according to the question the object has fallen to the half of its height (initial).

$\sf \implies$d=$\dfrac{h}{2}=\dfrac{5}{2}\:m$

➝The P.E. at that height will be(at point B) .,

$\sf \implies P.E._{B}=mg(h-d)$

$\sf \implies P.E._{B}=40\times9.8\times(5-\frac{5}{2})$

$\sf \implies P.E._{B}=\cancel{40}\times9.8\times\frac{5}{\cancel{2}}$

$\sf \implies P.E._{B}=20\times9.8\times5$

$\sf \implies\boxed{\large{\red{ P.E._{B}\:=\:980 \: Joule}}}$

∴ The Potential Energy of the body at that position will be = 980 joule

➝ The K.E,at that height will be(at point B) .,

$\sf \implies K.E._{B}=mgd$

$\sf \implies K.E._{B}=\cancel{40}\times9.8\times\dfrac{5}{\cancel2}$

$\sf \implies K.E._{B}=20\times9.8\times5$

$\sf \implies\boxed{\large{\red{ K.E._{B}\:=\:980 \: Joule}}}$

∴ The kinetic Energy of thae body at that position will be = 980 joule

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