Work Power Energy Question 4
Question 4 - 29 January - Shift 1
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: (300)
Solution:
Formula: Forth equation of motion
Displacement is $8^{m}$ sec.
$S_8=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
$\mathbf{S} _{\mathbf{8}}=5 \times 15$
$\Delta U=0.4 \times 10 \times 5 \times 15$
$\Delta U=20 \times 15=300$
