JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(300\)
- B \(30\)
- C \(3\)
- D \(150\)
Answer & Solution
Correct Answer
(A) \(300\)
Step-by-step Solution
Detailed explanation
Displacement is \(8^{\text {th }} sec\). \(S _8=0+\frac{1}{2} \times 10 \times(2 \times 8-1)\) \(S _8=5 \times 15\) \(\Delta U =0.4 \times 10 \times 5 \times 15\) \(\Delta U =20 \times 15=300\)
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