Question

A body of mass 2 kg is dropped from a height of 1 m. Its kinetic energy as it touches the ground is (g =10 metre per second square)​

Answer 1

When the body touches the ground

It's final velocity =v

It's initial velocity=u=0

m=2Kg

Now

$v^2 = u^2+2as$

$v^2= 2gs => v^2= 20 \\\\ KE= \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 20 J => 20 Joules$

Answer 2

Given :

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• $\sf Mass = 2 kg$
• $\sf Height = 1 m$
• $\sf g = 10 m/s$

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To find :

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• $\sf Kinetic\: Energy = ??$

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Solution :

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Acc. to law of conservation of energy :-⠀

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$\bigstar\:{\underline{\sf Total \: Potential \: Energy \: = \: Total \: kinetic \: energy }}$

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$\therefore \star\;{\boxed{\sf{\pink{Potential \: energy = Mass \times Gravity \times \: Height }}}}$

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$:\implies\sf Potential \: Energy = 2kg \times 10 m/s \times 1m$

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$:\implies\sf Potential \: Energy = 20 kg m^2 s^{-1}$

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$:\implies{\underline{\boxed{\frak{\purple{Potential \: Energy = 20 joules }}}}}\;\bigstar$

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$\sf Total \: Kinetic \: energy \: = \: Total \: potential \: energy$

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$\sf Total \: Kinetic \: Energy \: = 20 joules$

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$:\implies{\underline{\boxed{\frak{\purple { K. E. \: = 20 joules }}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Kinetic \;energy \; of \;the \: body \; when \; it \; touches \: the \: ground \: is\: \bf{20\: Joules }.}}}$

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