# OLDHW14 - oldhomewk 14 PAPAGEORGE MATT Due 4:00 am Question...

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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 = m g 3 T 1 = m g 3 k 1 x 1 = m g x 1 = m g 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 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 1 + m 2 ) g s - 1 2 k s 2 - μ m 1 g s . 3. K = m 2 g s - 1 2 k s 2 - μ ( m 1 + m 2 ) g s . 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 2 g s + 1 2 k s 2 + μ m 1 g s . 7. K = - ( m 1 + m 2 ) g s + 1 2 k s 2 + μ m 1 g s .

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oldhomewk 14 – PAPAGEORGE, MATT – Due: Feb 19 2008, 4:00 am 2 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 m 1 plus that of m 2 at B is given by K = K B = ( U g A - U g B ) + ( U sp A - U sp B ) - W dis
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