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Q.

A 0.4 kg mass takes 8s to reach ground when dropped from a certain height ‘P’ above surface of earth. The loss of potential energy in the last second of fall is ________J. (Take g = 10m/s²)

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answer is 300.

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Detailed Solution

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Hence,
$\begin{array}{l} h = \frac{1}{2} {gt}^{2} \\ = 320 m \end{array}$
Distance travelled in last second (8th seconds)
$\begin{array}{l} S_{8^{th}} = u + \frac{a}{2} (2 n - 1) \\ = 0 + \frac{10}{2} (2 \times 8 - 1) \\ = 5 \times 15 = 75 m \\ ∴ Δ \cup = {mgS}_{8^{th}} \\ = 0.4 \times 10 \times 75 \\ = 300 J \end{array}$

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