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# 13 (3) - hernandez(ejh742 oldhomework 13 Turner(58120 This...

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hernandez (ejh742) – oldhomework 13 – Turner – (58120) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The pulley system is in equilibrium, and the pulleys are weightless and frictionless. The spring constant is 2 N / cm and the suspended mass is 17 kg. 17 kg 2 N / cm How much will the spring stretch? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 27 . 7667 cm. Explanation: Let : k 1 = 2 N / cm , m = 17 kg , and g = 9 . 8 m / s 2 . m 1 k 1 T 1 T 1 T 1 T 2 The existence of a spring in a string defines the tension in the string because the force (tension) exerted by a spring is T = F = k x . At any point in the system summationdisplay F up = summationdisplay F down . At pulley 1, T 2 = 2 T 1 . At the suspended mass, T 2 + T 1 = m g 3 T 1 = m g 3 k 1 x 1 = m g x 1 = m g 3 k 1 = (17 kg) ( 9 . 8 m / s 2 ) 3 (2 N / cm) = 27 . 7667 cm . 002 (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 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 2 g s - 1 2 k s 2 - μ ( m 1 + m 2 ) g s . 2. K = - m 2 g s + 1 2 k s 2 + μ ( m 1 + m 2 ) g s . 3. K = m 2 g s - 1 2 k s 2 - μ m 1 g s . correct 4. K = ( m 1 + m 2 ) g s + 1 2 k s 2 - μ m 1 g s . 5. K = - ( m 1 + m 2 ) g s + 1 2 k s 2 + μ m 1 g s . 6. K = ( m 1 + 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 g s .

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hernandez (ejh742) – oldhomework 13 – Turner – (58120) 2 8. K = ( m 1 + m 2 ) g s - 1 2 k s 2 - μ m 1 g s . 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
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13 (3) - hernandez(ejh742 oldhomework 13 Turner(58120 This...

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