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PHY303K Test2

PHY303K Test2 - Version 014 TEST02 TSOI(58160 This...

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Version 014 – TEST02 – TSOI – (58160) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An ore car of mass 37000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 28 m lower vertically, is a horizontally situated spring with constant 3 . 3 × 10 5 N / m. The acceleration of gravity is 9 . 8 m / s 2 . Ignore friction. How much is the spring compressed in stop- ping the ore car? 1. 9.86343 2. 6.7462 3. 8.05025 4. 5.34743 5. 5.55889 6. 7.94607 7. 5.00213 8. 5.0898 9. 7.84424 10. 5.63231 Correct answer: 7 . 84424 m. Explanation: Energy is conserved, so the change of po- tential energy from when the car is at rest to when it just hits the spring is m g h = 1 2 m v 2 . The kinetic energy is then converted to po- tential energy in the spring as the cart comes to rest. When the spring is fully compressed by an amount d , all of the kinetic energy has been converted to potential energy so 1 2 m v 2 = 1 2 k d 2 . Thus, 1 2 k d 2 = m g h , and solving for d we have d = radicalbigg 2 m g h k = radicalBigg 2 (37000 kg) (9 . 8 m / s 2 ) (28 m) (3 . 3 × 10 5 N / m) = 7 . 84424 m . 002 10.0 points According to some nineteenth-century geo- logical theories (now largely discredited), the Earth has been shrinking as it gradually cools. If so, how would g have changed over geo- logical time? 1. It would not change; the mass of the Earth remained the same. 2. It would decrease; the Earth’s radius is decreasing. 3. It would increase; g is inversely propor- tional to the square of the radius of the Earth. correct Explanation: The acceleration g 1 r 2 of gravity would increase if the Earth shrank but its mass stayed the same. 003 (part 1 of 2) 10.0 points The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The block of mass m 1 lies on a rough horizontal surface with a constant coefficient of kinetic friction μ . This block is connected to a spring with spring constant k . The second block has a mass m 2 . The system is released from rest when the spring is unstretched, and m 2 falls a distance h before it reaches the lowest point. Note: When m 2 is at the lowest point, its velocity is zero. m 1 m 2 k m 1 m 2 h h μ

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Version 014 – TEST02 – TSOI – (58160) 2 Consider the moment when m 2 has de- scended by a distance s , where s is less than h . At this moment the sum of the kinetic energy for the two blocks K is given by 1. K = ( m 1 + m 2 ) g s + 1 2 k s 2 - μ m 1 g s . 2. K = ( m 1 + m 2 ) g s - 1 2 k s 2 + μ m 1 g s . 3. K = ( m 1 + m 2 ) g s - 1 2 k s 2 - μ m 1 g s . 4. K = - ( m 1 + m 2 ) g s + 1 2 k s 2 + μ m 1 g s . 5. K = - m 2 g s + 1 2 k s 2 + μ ( m 1 + m 2 ) g s . 6. K = - m 2 g s + 1 2 k s 2 + μ m 1 g s . 7. K = m 2 g s - 1 2 k s 2 - μ ( m 1 + m 2 ) g s . 8. K = m 2 g s - 1 2 k s 2 - μ m 1 g s . correct Explanation: Basic Concepts: Work-Energy Theorem Spring Potential Energy Frictional Force according to the Work- Energy Theorem Solution: W ext A B = ( K B - K A ) + ( U g B - U g A ) + ( U sp B - U sp A ) + W dis A B For the present case, the external work W ext A B = 0, A corresponds to the initial state and B the state where m 2 has descended by a distance s . The sum of the kinetic energy of
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PHY303K Test2 - Version 014 TEST02 TSOI(58160 This...

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