October 10 Solutions

# October 10 Solutions - 7.33 IDENTIFY Apply Eq(7.16 SET UP...

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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 F x x dx       . 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. 7.36. IDENTIFY: Apply Eq.(7.18). SET UP: 23 12 d dx x x     and d dy y y . EXECUTE: ˆˆ UU xy   F = i j since U has no z -dependence. 22 and so ,   3 3 3 3 2 x y x y  ij F = i + j +  . EVALUATE: x F and x have the same sign and y F and y have the same sign. When 0 x , x F is in the -direction, x and so forth. 7.83. 2 ˆ , Fj 3 2.50 N/m IDENTIFY: F is not constant so use Eq.(6.14) to calculate W . F must be evaluated along the path. (a) SET UP: The path is sketched in Figure 7.83a. d dx dy  l i j 2 d xy dy   Fl On the path, so 3 d y dy   Figure 7.83a EXECUTE:   2 2 1 1 2 3 4 4 4 21 1 ( )

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October 10 Solutions - 7.33 IDENTIFY Apply Eq(7.16 SET UP...

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