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Unformatted text preview: lenoir (wml297) oldhomework 13 Turner (58220) 1 This printout should have 10 questions. Multiplechoice 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 9 N / cm and the suspended mass is 16 kg. 16 kg 9 N / cm How much will the spring stretch? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 5 . 80741 cm. Explanation: Let : k 1 = 9 N / cm , m = 16 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 = mg 3 T 1 = mg 3 k 1 x 1 = mg x 1 = mg 3 k 1 = (16 kg) ( 9 . 8 m / s 2 ) 3 (9 N / cm) = 5 . 80741 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 . 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 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 + ( m 1 + m 2 ) g s. 2. K = ( m 1 + m 2 ) g s 1 2 k s 2 + m 1 g s. 3. K = m 2 g s + 1 2 k s 2 + m 1 g s. 4. K = m 2 g s 1 2 k s 2 ( m 1 + m 2 ) g s. 5. K = ( m 1 + m 2 ) g s 1 2 k s 2 m 1 g s. 6. K = ( m 1 + m 2 ) g s + 1 2 k s 2 + m 1 g s. 7. K = ( m 1 + m 2 ) g s + 1 2 k s 2 m 1 g s. lenoir (wml297) oldhomework 13 Turner (58220) 2 8. K = m 2 g s 1 2 k s 2 m 1 g s. correct Explanation: Basic Concepts: WorkEnergy 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...
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This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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