Work Energy Power

JEE Main level, numerical answer, medium

76 of 183

Question 76 of 183

A 0.4kg0.4kg mass takes 8s8s to reach ground when dropped from a certain height PP above surface of earth. The loss of potential energy in the last second of fall is ______ JJ. [Take g=10ms−2g=10m{s}^{-2}]

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Answer: 300

Displacement in nth{n}^{th} second is given by Sn=u+a2(2n−1){S}_{n}=u+\dfrac{a}{2}\left(2n-1\right)
Here, displacement is 8th second is
S8=0+12×10×(2×8−1){S}_{8}=0+\dfrac{1}{2}\times 10\times (2\times 8-1)
⇒S8=75m\Rightarrow {S}_{8}=75m
Loss of potential energy is
ΔU=mgS8=0.4×10×75\Delta U=mg{S}_{8} =0.4\times 10\times 75
⇒ΔU=300J\Rightarrow \Delta U=300J.

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