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**Unformatted text preview: **homework 17 – HETTIGER, CHRISTOF – Due: Feb 28 2008, 4:00 am 1 Question 1, chap 8, sect 2. part 1 of 1 10 points The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 5 kg block lies on a rough horizontal surface with a constant coefficient of kinetic friction 0 . 1. This block is connected to a spring with spring constant 2 N / m. The second block has a mass of 7 kg. The system is released from rest when the spring is unstretched, and the 7 kg block falls a distance h before it reaches the lowest point. The acceleration of gravity is 9 . 8 m / s 2 . Note: When the 7 kg block is at the lowest point, its velocity is zero. 5 kg 7 kg 2 N / m 5 kg 7 kg h h μ = 0 . 1 Calculate the mechanical energy removed by friction durning the time when the 7 kg mass falls a distance h. Correct answer: 312 . 13 J (tolerance ± 1 %). Explanation: Basic Concepts: Work-Energy Theorem Spring Potential Energy Frictional Force according to the Work- Energy Theorem m 1 m 2 k m 1 m 2 h h μ Given : m 1 = 5 kg , m 2 = 7 kg , μ = 0 . 1 , and k = 2 N / m . 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 A → B = m 2 g s − 1 2 k s 2 − μm 1 g s. (1) Based on the Eq. 1 at s = h , K B = 0, we have μm 1 g h = m 2 g h − 1 2 k h 2 . In turn, h = 2 g [ m 2 − μm 1 ] k (2) = 2 (9 . 8 m / s 2 ) [(7 kg) − (0 . 1) (5 kg)] (2 N / m) = 63 . 7 m . We know that E initial = E final + E μ , where E μ is the mechanical energy removed by fric- tion. In order to solve the second part of the problem we need to calculate the initial and fi- nal energies. Let y 1 be the vertical position of m 1 and y 2 the vertical position of...

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