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/s$^2$)
Answer (integer)
300
Solution
Displacement is $8^{\text {th }}$ sec.
<br/><br/>
$\mathrm{S}_{8}=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
<br/><br/>
$\mathbf{S}_{8}=5 \times 15$
<br/><br/>
$\Delta \mathrm{U}=0.4 \times 10 \times 5 \times 15$
<br/><br/>
$\Delta \mathrm{U}=20 \times 15=300$
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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