221_PartUniversity Physics Solution

221_PartUniversity Physics Solution - 7-10 7.30. Chapter 7...

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7-10 Chapter 7 7.30. IDENTIFY and SET UP: The friction force is constant during each displacement and Eq.(6.2) can be used to calculate work, but the direction of the friction force can be different for different displacements. 2 k (0.25)(1.5 kg)(9.80 m/s ) 3.675 N; fm g μ == = direction of f ! is opposite to the motion. EXECUTE: (a) The path of the book is sketched in Figure 7.30a. Figure 7.30a For the motion from you to Beth the friction force is directed opposite to the displacement s ! and 1 (3.675 N)(8.0 m) 29.4 J. Wf s =− For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and 2 29.4 J. W tot 1 2 29.4 J 29.4 J 59 J WW W =+= = (b) The path of the book is sketched in Figure 7.30b. 2 2(8.0 m) 11.3 m s Figure 7.30b f ! is opposite to , s ! so (3.675 N)(11.3 m) 42 J s (c) For the motion from Kim to you 29.4 J s = −= Figure 7.30d The total work for the round trip is 29.4 J 29.4 J 59 J. −− = (d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero. Both these results show that the friction force is nonconservative. 7.31. IDENTIFY: The work done by a spring on an object attached to its end when the object moves from i x to f x is 22 11 if Wk x k x . This result holds for any i x and f x . SET UP: Assume for simplicity that 1 x , 2 x and 3 x are all positive, corresponding to the spring being stretched. EXECUTE: (a) 1 12 2 () kx x (b) 1 2 . The total work is zero; the spring force is conservative. (c) From 1 x to 3 , x 1 31 2 . x x From 3 x to 2 x , 1 23 2 . x x The net work is 1 21 2 . This is the same as the result of part (a). EVALUATE: The results of part (c) illustrate that the work done by a conservative force is path independent. For the motion from you to Kim s = − (3.675 N)(8.0 m) 29.4 J W = Figure 7.30c
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Potential Energy and Energy Conservation 7-11 7.32. IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from ( ). Ux Use coordinates where the origin is at one atom. The other atom then has coordinate x . EXECUTE: 66 6 7 16 x dU d C d C FC dx dx x dx x x ⎛⎞ =− =+ ⎜⎟ ⎝⎠ The minus sign mean that x F is directed in the -direction, x toward the origin. The force has magnitude 7 6 6/ Cx and is attractive. EVALUATE: U depends only on x so F ! is along the x -axis; it has no y or z components. 7.33. IDENTIFY: Apply Eq.(7.16). SET UP: The sign of x F indicates its direction. EXECUTE: 4 33 4( 4 . 8 J m ) x dU Fx x dx α . 4 3 ( 0.800 m) (4.8 J m )( 0.80 m) 2.46 N. x F −= The force is in the -direction. x + EVALUATE: 0 x F > when 0 x < and 0 x F < when 0 x > , so the force is always directed towards the origin. 7.34. IDENTIFY: Apply () dU x dx . SET UP: 2 (1/ ) 1 dx dx x EXECUTE: 12 2 (/ ) ( 1 / ) x d Gmm x d x Gmm G mm dx dx x ⎡⎤ = ⎢⎥ ⎣⎦ . The force on 2 m is in the -direction x . This is toward 1 m , so the force is attractive.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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221_PartUniversity Physics Solution - 7-10 7.30. Chapter 7...

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