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JEE Main 2023 (29 Jan, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2023 (29 Jan, Shift 1)

Q17JEE Main 2023 · 29 JAN · SHIFT 1Medium

Conservation of Energy

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 = 10 m/s2{}^{2})

Numerical answer type. Work it out, then open the answer.

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Answer: 300

  1. Distance fallen in the nnth second from rest: sn=g2(2n−1)s_n = \frac{g}{2}(2n - 1).
  2. s8=5×15=75 ms_8 = 5 \times 15 = 75\text{ m}
  3. Loss of PE =mgs8=0.4×10×75= mgs_8 = 0.4 \times 10 \times 75
  4. =300 J= 300\text{ J}
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