# oldhw15 - hinojosa(jlh3938 – oldhomework 15 – Turner...

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Unformatted text preview: hinojosa (jlh3938) – oldhomework 15 – Turner – (58185) 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 16 kg. 16 kg 2 N / cm How much will the spring stretch? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 26 . 1333 cm. Explanation: Let : k 1 = 2 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 (2 N / cm) = 26 . 1333 cm . keywords: 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 1 + m 2 ) g s- 1 2 k s 2 + μm 1 g s. 2. K =- m 2 g s + 1 2 k s 2 + μ ( m 1 + m 2 ) g s. 3. K = ( m 1 + m 2 ) g s- 1 2 k s 2- μm 1 g s. 4. K =- m 2 g s + 1 2 k s 2 + μm 1 g s. 5. K = m 2 g s- 1 2 k s 2- μ ( m 1 + m 2 ) g s. hinojosa (jlh3938) – oldhomework 15 – Turner – (58185) 2 6. K = m 2 g s- 1 2 k s 2- μm 1 g s. correct 7. K = ( m 1 + m 2 ) g s + 1 2 k s 2- μm 1 g s. 8. K =- ( m 1 + m 2 ) g s + 1 2 k s 2 + μm 1 g s....
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oldhw15 - hinojosa(jlh3938 – oldhomework 15 – Turner...

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