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Unformatted text preview: oldhomewk 14 PAPAGEORGE, MATT Due: Feb 19 2008, 4:00 am 1 Question 1, chap 5, sect 5. part 1 of 1 10 points The pulley system is in equilibrium, and the pulleys are weightless and frictionless. The spring constant is 8 N / cm and the suspended mass is 12 kg. The acceleration of gravity is 9 . 8 m / s 2 . 12 kg 8 N / cm How much will the spring stretch? Correct answer: 4 . 9 cm (tolerance 1 %). Explanation: Let : k 1 = 8 N / cm , m = 12 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 = (12 kg) ( 9 . 8 m / s 2 ) 3 (8 N / cm) = 4 . 9 cm . Question 2, chap 8, sect 1. part 1 of 2 10 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 + m 2 ) g s. 4. K = ( m 1 + m 2 ) g s 1 2 k s 2 + m 1 g s. 5. K = ( m 1 + m 2 ) g s + 1 2 k s 2 m 1 g s. 6. K = 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. oldhomewk 14 PAPAGEORGE, MATT Due: Feb 19 2008, 4:00 am 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 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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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
 Physics, Friction, Mass

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