JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 23)
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$$^2$$)
Одговорити
300
Објашњење
Displacement is $8^{\text {th }}$ sec.
$\mathrm{S}_{8}=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
$\mathbf{S}_{8}=5 \times 15$
$\Delta \mathrm{U}=0.4 \times 10 \times 5 \times 15$
$\Delta \mathrm{U}=20 \times 15=300$
$\mathrm{S}_{8}=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
$\mathbf{S}_{8}=5 \times 15$
$\Delta \mathrm{U}=0.4 \times 10 \times 5 \times 15$
$\Delta \mathrm{U}=20 \times 15=300$