answerkey6 - In other words, all kinetic energy is...

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Answer Key for problems in Unit VI. 3. m = 5kg, h = 2 . 5m P . E . = mgh = 5kg × 9 . 8m / s 2 × 2 . 5m = 122 . 5J. 5. m = 600kg, v = 25m / s K . E . = 1 2 mv 2 = 1 2 600kg(25m / s) 2 = 187500J 8. W = ~ F · ~ d = 185N × 12m = 2220N, P = W/t = 2220N / 10s = 220W. 9. The speed will be the maximum at the lowest point. The acceleration will be the smallest at the flat path. 10. Work P . E . = mgh W mat = 90 × 9 . 8 × 10 = 8820J P mat = 8820J / 7 . 5s = 1176W W joe = 70 × 9 . 8 × 10 = 6860J P mat = 6860J / 5 . 5s = 1247 . 27W 12. Free fall motion a = g = 9 . 8m / s. (a) d = 1 2 gt 2 t = s 2 d g = s 2 × 75m 9 . 8m / s 2 = 3 . 9123s (b) v = at = gt = 9 . 8m / s 2 × 3 . 9123s = 38 . 3406m / s (c) Potential Energy = mgh = 75kg9 . 8m / s 2 75m = 55125J (d) 1 2 mv 2 = mgh 1
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v 2 = 2 gh v = q 2 gh = 38 . 3406m / s (e) We don’t need mass because mass terms are canceled in the equation of energy conservation. 13. At the maximum height, the velocity will be zero.
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Unformatted text preview: In other words, all kinetic energy is transformed to the potential energy. K . E . = 10J = mgh h = 10J mg = 10J . 2kg 9 . 8m / s 2 = 5 . 10204m 14. (a) P . E . = mgh = 1 . 5kg 9 . 8m / s 2 100m = 1470J . (b) K . E . = 1 2 mv 2 = 1 2 1 . 5kg (12m / s) 2 = 108J (c) Total Energy = P.E. + K.E. = 1578J (d) Just before it reaches the ground, all potential energy is transformed to kinetic energy. Total Energy = 1 2 mv 2 v = s 2 Total Energy m = s 2 1578J 1 . 5kg = 45 . 8694m / s 2...
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answerkey6 - In other words, all kinetic energy is...

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