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  • 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 J) in the last second of fall is ? <br

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 J) in the last second of fall is ? 
(Take g = 10 m/s2)

Option: 1

300


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ \begin{aligned} & \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & \mathrm{~h}=0+\frac{1}{2} \cdot \mathrm{g}(8)^2=\frac{10}{2} \times 8 \times 8=320 \mathrm{~m} \end{aligned}

Distance covered in last second

\begin{aligned} & h_1=u+\frac{a}{2}(2 n-1) \\ & =0+\frac{10}{2}[2(8)-1] \\ & h_i=5[15]=75 \mathrm{~m} \\ & \Delta U_{\text {loss }}=m g \Delta h \\ & \Delta U_{\text {loss }}=0.4 \times 10 \times 75=300 \mathrm{~J} \\ & \text { Ans } \rightarrow 300 \mathrm{~J} \end{aligned}

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shivangi.shekhar

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