A body of mass 6 g falls under action of gravity. At initial position 'A' its P.E. is 480 J and K.E. is 0 J. During its downward journey at point 'B' its energies will be (g = 10 ms-2)
Correct answer: A. P.E. = 300 J and K.E. = 180 J
- A. P.E. = 300 J and K.E. = 180 J
- B. P.E. = 180 J and K.E. = 300 J
- C. P.E. = 230 J and K.E. = 250 J
- D. P.E. = 250 J and K.E. = 230 J
Explanation
Explanation:At point A, P.E = 480 JSince, P.E = mgh 480 = 6 × 10 × h h = 480 / 60 h = 8m above groundAt B, having fallen 3m, height above ground at B = 8 - 3 = 5mP.E at B = mgh = 6 × 10 × 5 = 300 JK.E at B = 480 - 300 = 180 J as total energy is conserved at any point.
This question appeared on the UHS MDCAT 2009 paper, which you can sit online with every answer explained.
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