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Unformatted text preview: A roller coaster cart oF mass m = 379 kg starts stationary at point A, where h 1 = 21 . 3 m and a while later is at B , where h 2 = 10 . 5 m. The acceleration oF gravity is 9 . 8 m / s 2 . h h A B 1 2 What is the potential energy oF the cart relative to the ground at A ? Correct answer: 79112 . 5 J. Explanation: The potential energy U at a is given by U = mg h 1 = (379 kg) (9 . 8 m / s 2 ) (21 . 3 m) = 79112 . 5 J . 003 (part 2 oF 2) 10.0 points What is the speed oF the cart at B , ignoring the eect oF Friction? Correct answer: 14 . 5492 m / s. Explanation: By conservation oF energy U + K = 0 ( mg h 2mg h 1 ) + 1 2 mv 2 = 0 mg ( h 1h 2 ) = 1 2 mv 2 Solving For v , v = r 2 g ( h 1h 2 ) = R 2 (9 . 8 m / s 2 ) (21 . 3 m10 . 5 m) = 14 . 5492 m / s ....
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 Fall '09
 MIHALISIN
 Physics, Energy

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