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Exam2-solutions - Version 040 Exam2 markert(56480 This...

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Version 040 – Exam2 – markert – (56480) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rock of mass m is thrown horizontally off a building from a height h . The speed of the rock as it leaves the thrower’s hand at the edge of the building is v 0 , as shown. m h v 0 m What is the kinetic energy of the rock just before it hits the ground? 1. K f = 1 2 m v 2 0 + m g h correct 2. K f = m g h - 1 2 m v 2 0 3. K f = m g h 4. K f = 1 2 m v 2 0 - m g h 5. K f = 1 2 m v 2 0 Explanation: We can use work-energy theorem, K f - K i = W , where in the present problem, W = ( m g ) h , and K i = 1 2 m v 2 0 . Thus, K f = K i + W = 1 2 m v 2 0 + m g h . Comments for Matter Interaction readers: The energy principle in MI-text is referred to as the work-energy theorem in traditional text. Here the work done by the gravitational force near the surface of the earth is given by: W = ( - m g )( - h ) = m g h . 002 10.0 points If 3.4 J of work is done in raising a 177 g apple, how far is it lifted? 1. 2.20315 2. 1.68135 3. 1.47914 4. 1.95611 5. 1.71018 6. 1.13896 7. 1.60514 8. 1.567 9. 2.12324 10. 1.19835 Correct answer: 1 . 95611 m. Explanation: Basic Concepts: W applied = F applied d cos θ = mgd since θ = 0 cos θ = 1. 1 J = 1 N · m F = mg 1 N = 1 kg · m / s 2 Given: W applied = 3 . 4 J m = 177 g g = 9 . 81 m / s 2 Solution: The force and displacement are parallel, so the distance is given by d = W applied mg = 3 . 4 J (177 g)(9 . 82 m / s 2 ) · 1000 g 1 kg = 1 . 95611 m 003 10.0 points A ball rolls up an incline, slows to a stop and rolls back down. direction of motion A B C D E F G
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Version 040 – Exam2 – markert – (56480) 2 Which of the following represents the di- rection of the ball’s acceleration at the times indicated? 1. A B C D E F G 2. A B C D E F G correct 3. A B C D E F G 4. A B C D E F G 5. A B C D E F G 6. A B C D E F G 7. A B C D E F G Explanation: vectora = Δ vectorv Δ t , so vectora is in the same direction as Δ vectorv . By subtracting the velocity vectors, one can see that the direction of Δ vectorv is exactly opposite the direction of vectorv . Therefore, the direction of vectora is down the incline at all points.
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