chap 7 - CONSERVATION OF ENERGY 7 EXERCISES Section 7.1...

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7.1 CONSERVATION OF ENERGY E XERCISES Section 7.1 Conservative and Nonconservative Forces 11. I NTERPRET In this problem we want to find the work done by the frictional force in moving a block from one point to another over two different paths. D EVELOP Figure 7.16 is a plan view of the horizontal surface over which the block is moved, showing the paths (a) and (b) . The force of friction is f n mg = = µ µ opposite to the displacement. Since f is constant, using Equation 6.1, the work done is W f r f r = = − E VALUATE The work done by friction along path (a) with r L L L a = + = 2 is W f r mg L mgL a a = − = − = − µ µ ( ) 2 2 Similarly, the work done by friction along path (b) with r L L L b = + = 2 2 2 is W f r mg L mgL b b = − = − = − µ µ ( ) 2 2 A SSESS Since the work done depends on the path chosen, friction is not a conservative force. 12. Take the origin at point 1 in Fig. 7.16 with the x axis horizontal to the right and the y axis vertical upward. The gravitational force on an object is constant, F mg j g = − ˆ , while the paths are (a) dr j dy = ˆ for x = 0 and 0 y L , followed by dr idx = ˆ for y L = and 0 x L , and (b) dr idx jdy i j dy = = ˆ ˆ ( ˆ ˆ ) , + + for 0 y L (since x y = along this path). The work done by gravity (Equation 6.11) is W F dr mg j jdy mgj ndx g a g L ( ) ( ˆ ) ˆ ( ˆ ) = = + 0 0 0 0 L L mg dy mgL = − + = − and W mg j n j dy mg dy mgL g b L ( ) ( ˆ ) ( ˆ ) = + = − = − 0 0 Of course, these must be the same because gravity is a conservative force. Section 7.2 Potential Energy 13. I NTERPRET The problem is about gravitational potential energy relative to a reference point of zero energy. In Example 7.1, the reference point was taken to be the 33rd floor. In this problem, we take the street level to be our reference point. D EVELOP The change in potential energy with a change in the vertical distance y is given by Equation 7.3, U mg y = . Each floor is 3.5 m high. E VALUATE (a) The office of the engineer is on the 33rd floor, or is 32 stories above the street level (the first floor) where U 1 0 = . Thus, the difference in gravitational potential energy is U U U U mg y = = = = × × = 33 1 33 0 55 9 8 32 3 5 N m ( . )( . ) 60 4 . kJ (b) At the 59th floor, U U U = = × × = 59 1 55 9 8 58 3 5 109 ( . )( . ) . N m kJ (c) Street level is the zero of potential energy, U 1 0 = . 7
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7.2 Chapter 7 A SSESS Potential energy depends on the reference point chosen, but potential energy difference between two points does not. What matters physically is the difference in potential energy. The differences in potential energy between any two levels are the same as in Example 7.1, e.g., U U 59 33 109 60 4 = = ( . ) kJ kJ 49.0 kJ. 14. If we define the zero of potential energy to be at zero altitude ( ), y = 0 then U ( ) , 0 0 = and Equation 7.3 (for the gravitational potential energy near the surface of the Earth, | | << km y 6370 ) gives U y U U y mg y ( ) ( ) ( ) ( ) = = = 0 0 mgy . Therefore, (a) U ( ) ( . )( ) . , 1900 70 9 8 1900 1 30 m N m MJ = × = and (b) U ( ) ( . )( ) = × = 86 70 9 8 86 m N m 59 0 . . kJ 15. I NTERPRET The problem is about the change in gravitational potential energy as a person comes down from the mountaintop.
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