Question

A body of mass 10.5 kg is dropped from the top of a building of height 9 m.
A. What Force acting on the body during its fall?
B. Change in potential energy?
C. Kinetic energy of the body just before striking the ground? [g-9.8m/s

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Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the concept of Second Law of Motion, Potential Energy and Kinetic Energy has been used. We are already given mass and acceleration. Using that we can find out the force exerted on the body. The by using the formula of Potential Energy we can find initial potential energy. Then, we can find the final potential by using height as 0 since height just before ground will be zero only. Then we can subtract them to find change in Potential Energy. Then we can find out Kinetic Energy.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Force,\;(F)\;=\;\bf{Mass\;(m)\;\times\;Acceleration\;(a)}}}$

$\\\;\boxed{\sf{Potential\;Energy\;=\;\bf{mgh}}}$

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★ Solution :-

Given,

» Mass of the body = m = 10.5 Kg

» Height of the building = h = 9 m

» Acceleration on body (gravity) = a = 9.8 m/sec²

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A.) For Force experienced by the body ::

$\\\;\;\;\sf{:\rightarrow\;\;Force,\;(F)\;=\;\bf{Mass\;(m)\;\times\;Acceleration\;(a)}}$

$\\\;\;\;\sf{:\rightarrow\;\;Force_{(Ball)},\;(F)\;=\;\bf{10.5\;\times\;9.8}}$

$\\\;\;\;\sf{:\rightarrow\;\;Force_{(Ball)},\;(F)\;=\;\bf{10.5\;\times\;9.8}}$

$\\\;\;\;\bf{:\rightarrow\;\;Force_{(Ball)},\;(F)\;=\;\bf{102.9\;\;N}}$

This is because, Kg m/sec² = N

$\\\;\underline{\boxed{\tt{Hence,\;\;Force\;\;experienced\;\;by\;\;ball\;=\;\bf{102.9\;\;N}}}}$

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B.) For the change in Potential Energy of the body ::

~ First let's find Initial Potential Energy -

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy\;=\;\bf{mgh}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\bf{10.5\;\times\;9.8\;\times\;9}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\bf{10.5\;\times\;9.8\;\times\;9}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy_{(Initial)}\;=\;\underline{\underline{\bf{926.1\;\;Joules}}}}$

This is because, Kg m²/sec² = Joules.

~ Now let's find the Final Potential Energy of the ball -

Since, final potential energy is at ground. So at ground the height of the body will be 0. So,

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy\;=\;\bf{mgh}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\bf{10.5\;\times\;9.8\;\times\;0}}$

$\\\;\;\;\sf{:\Rightarrow\;\;Potential\;Energy_{(Final)}\;=\;\underline{\underline{\bf{0\;\;Joules}}}}$

~ Now change in Potential Energy is given as,

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

$\\\;\sf{:\Longrightarrow\;\;Change\;in\;Potential\;Energy\;=\;\bf{926.1\;Joules\;+\;0\;Joules}}$

$\\\;\bf{:\Longrightarrow\;\;Change\;in\;Potential\;Energy\;=\;\bf{926.1\;\;Joules}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;change\;\;in\;\;Potential\;\;Energy\;=\;\bf{926.1\;\;Joules}}}}$

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C.) For the Kinetic Energy of the ball just before it strikes the ground.

According to the Law of Conservation of Energy, total energy of the system remains constant.

Also when ball comes on ground, it is not given that it rebounds. So its velocity at ground will be zero thus, its kinetic energy will also be zero.

From this we can figure out that the highest kinetic energy of the body will be just before the ground.

Also, Kinetic energy of the body just before it strikes the ground will be equal to the potential energy of the body at height 9 m.

$\\\;\bf{:\mapsto\;\;Kinetic\;Energy\;of\;Ball\;=\;\bf{926.1\;\;Joules}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;kinetic\;\;energy\;\;of\;\;ball\;\;=\;\bf{926.1\;\;Joules}}}}$

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★ More Formulas to Know :-

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

$\\\;\sf{\leadsto\;\;\;Work\;=\;\vec{F}\;.\;\vec{s}\;\cos \theta}$

$\\\;\sf{\leadsto\;\;\;Change\;in\;Kinetic\;Energy\;=\;\dfrac{1}{2}\:mv^{2}\;-\;\dfrac{1}{2}\:mu^{2}}$

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

$\\\;\sf{\leadsto\;\;\;m_{1}\:u_{1}\;+\;m_{2}\:u_{2}\;=\;m_{1}\:v_{1}\;+\;m_{2}\:u_{2}}$

This ↑ is for for Elastic Motion in One Dimension.

Answer 2

$\huge\bold{\underbrace{Answer!}}$

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Given,

• Mass of the body = 10.5 kg
• Height of the building = 9m

To find,

• What Force acting on the body during its fall?
• Change in potential energy?
• Kinetic energy of the body just before striking the ground? [g-9.8m/s²]

Formula used,

• Force = Mass × Acceleration
• Potential energy = mgh
• Change in potential energy = Initial potential energy + Final potential energy

Solution,

Firstly, let's find out the force acting on the body during it's fall.

A) Force = Mass × Acceleration (Gravitation is pulling the body, so acceleration = 9.8m/s²)

${\implies}$ = 10.5 × 9.8

${\implies}$ = 102.9 Kgm/s²

Therefore, 102.9 kgm/s² is the force acting on the bodu during fall.

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Secondly, let's find the change in potential energy.

So, to find the change in potential energy, we should find initial and final potential energy.

B) Initial Potential energy = mgh

${\implies}$ = 10.5 × 9.8 × 9

${\implies}$ = 926.1 Joules

Final potential energy = mgh

${\implies}$ = 10.5 × 9.8 × 0

${\implies}$ = 0 joules

Change in potential energy = Initial potential energy + Final potential energy

${\implies}$ = 926.1 + 0

${\implies}$ = 926.1 joules

Therefore, 926.1 joules is the change in potential energy.

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Lastly, let's find the Kinetic energy of the body just before striking the ground.

C) According to the Law of conservation of energy, Potential enery of the body before falling and kinetic energy before stricking the ground is equal.

So, Potential energy = Kinetic energy

926.1 = 926.1

Therefore, Kinetic energy of the body just before striking the ground is 926.1 joules

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