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)

Question

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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Displacement is 8 text th sec. S 8=0+(1/2) × 10 ×(2 × 8-1) S 8=5 × 15 Δ U =0.4 × 10 × 5 × 15 Δ U =20 × 15=300

Q. A $0.4 k g$ 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 $g = 10 m / s^{2})$

Displacement is $8^{th}$ sec.
$S_{8} = 0 + \frac{1}{2} \times 10 \times (2 \times 8 - 1)$
$S_{8} = 5 \times 15$
$Δ U = 0.4 \times 10 \times 5 \times 15$
$Δ U = 20 \times 15 = 300$

Q. A 0.4kg 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=10m/s2)

Displacement is 8th sec. S8​=0+21​×10×(2×8−1) S8​=5×15 ΔU=0.4×10×5×15 ΔU=20×15=300

Q. A $0.4 k g$ 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 $g = 10 m / s^{2})$

Displacement is $8^{th}$ sec.
$S_{8} = 0 + \frac{1}{2} \times 10 \times (2 \times 8 - 1)$
$S_{8} = 5 \times 15$
$Δ U = 0.4 \times 10 \times 5 \times 15$
$Δ U = 20 \times 15 = 300$