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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online