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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2023
  • Medium
  • Numerical

A \(0.4\,kg\) mass takes \(8 s\) 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 \(g =10\,m / s ^2\) ]

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Answer & Solution

Correct Answer

(A) \(300\)

Step-by-step Solution

Detailed explanation

Displacement is \(8^{\text {th }} sec\).

\(S _8=0+\frac{1}{2} \times 10 \times(2 \times 8-1)\)

\(S _8=5 \times 15\)

\(\Delta U =0.4 \times 10 \times 5 \times 15\)

\(\Delta U =20 \times 15=300\)

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