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A body of mass $1 \mathrm{~kg}$ is thrown upwards with a velocity $20 \mathrm{~ms}^{-1}$. It momentarily comes to rest after attaining a height of $18 \mathrm{~m}$. How much energy is lost due to air friction ? $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$20 \mathrm{~J}$
Key Idea The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body posses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $=\mathrm{KE}-\mathrm{PE}$
$$
\begin{aligned}
& =\frac{1}{2} \mathrm{mv}^2-\mathrm{mgh} \\
& =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\
& =200-180=20 \mathrm{~J}
\end{aligned}
$$
Initially body posses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $=\mathrm{KE}-\mathrm{PE}$
$$
\begin{aligned}
& =\frac{1}{2} \mathrm{mv}^2-\mathrm{mgh} \\
& =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\
& =200-180=20 \mathrm{~J}
\end{aligned}
$$
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