cw12 - A roller coaster cart oF mass m = 379 kg starts...

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patel (kp9583) – CW12 – Mackie – (20211) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. HINT: For #3, conserve mechanical energy From A B. 001 10.0 points A car oF weight 1800 N operating at a rate oF 148 kW develops a maximum speed oF 36 m / s on a level, horizontal road. Assuming that the resistive Force (due to Friction and air resistance) remains constant, what is the car’s maximum speed on an incline oF 1 in 20; i.e. , iF θ is the angle oF the incline with the horizontal, sin θ = 1 / 20 ? Correct answer: 35 . 2288 m / s. Explanation: IF f is the resisting Force on a horizontal road, then the power P is P = f v horizontal , so that f = P v h = (1 . 48 × 10 5 W) (36 m / s) = 4111 . 11 N . On the incline, the resisting Force is F = f + mg sin θ = f + W 20 = P v h + W 20 . And, F v = P , so v = P F = P P v h + W 20 = (1 . 48 × 10 5 W) (1 . 48 × 10 5 W) (36 m / s) + (1800 N) 20 = 35 . 2288 m / s . 002 (part 1 oF 2) 10.0 points
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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 2-mg h 1 ) + 1 2 mv 2 = 0 mg ( h 1-h 2 ) = 1 2 mv 2 Solving For v , v = r 2 g ( h 1-h 2 ) = R 2 (9 . 8 m / s 2 ) (21 . 3 m-10 . 5 m) = 14 . 5492 m / s ....
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