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Unformatted text preview: midterm 02 – ALIBHAI, ZAHID – Due: Mar 7 2007, 11:00 pm 1 Question 1 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 3 kg block lies on a rough horizontal surface with a constant coefficient of kinetic friction 0 . 3. This block is connected to a spring with spring constant 7 N / m. The second block has a mass of 8 kg. The system is released from rest when the spring is unstretched, and the 8 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 8 kg block is at the lowest point, its velocity is zero. 3 kg 8 kg 7 N / m 3 kg 8 kg h h μ = 0 . 3 Calculate the mechanical energy removed by friction durning the time when the 8 kg mass falls a distance h. 1. 156 . 065 J 2. 162 . 068 J 3. 169 . 03 J 4. 175 . 342 J correct 5. 184 . 397 J 6. 190 . 543 J 7. 198 . 083 J 8. 207 . 446 J 9. 215 . 13 J 10. 221 . 852 J Explanation: Basic Concepts: WorkEnergy 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 = 3 kg , m 2 = 8 kg , μ = 0 . 3 , and k = 7 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 ) [(8 kg) − (0 . 3) (3 kg)] (7 N / m) = 19 . 88 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 m 2 , where say y = 0 is the initial vertical position of m 2 . The total energy of the system is E = U + K , midterm 02 – ALIBHAI, ZAHID – Due: Mar 7 2007, 11:00 pm 2 where U = U grav 1 + U grav 2 + U spring . Ini tially, U grav 1 ,in = m 1 g y 1 , U grav 2 ,in = 0, U spring i n = 0 (the spring is unstretched) and K in = 0 . In the final situation the masses have a null velocity, and so we have once again K fin = 0 . The potential energies are U grav 1 = m 1 g y 1 , U grav 2 ,fin = m 2 g ( y 1 − h ) and U spring = 1 2 k h 2 . Finally, letting y 1 = 0 m and using Eq. 2 for h , we get E μ = E initial − E final = − m 2 g ( − h ) − 1 2 k h 2 = m 2 g h − 1 2 k h 2 = 2 g 2 k bracketleftbig m 2 ( m 2 − μm 1 ) − ( m 2 − μm 1 ) 2 bracketrightbig = 2 μm 1 g 2 k [ m 2 − μm 1 ] = 2 (0 . 3) (3 kg) (9 . 8 m / s 2 ) 2 (7 N / m) × [(8 kg) − (0 . 3) (3 kg)]...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Friction, Mass, Light

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