T1a5s - Assignment 5 solutions(Fall term Physics 102 5.13(a...

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5.13 (a) We use the work-energy theorem to find the work. () 2 22 11 1 0 70 kg 4.0 m s 2 i WK Em v m v = =−= = 2 5.6 10 J −× . Assignment 5 solutions (Fall term Physics 102) (b) ( ) ( ) ( ) ( ) cos cos180 kk k sf s n s m g s θµ µ == ° = = WF , so ( ) ( ) 2 2 5.6 0.70 70 kg 9.80 m s k W s mg =− = 1.2 m . 5.25 (a) When the rope is horizontal, the swing is 2.0 m above the level of the bottom of the circular arc, and ( )( ) 40 N 2.0 m g PE mgy == = 80 J . (b) When the rope makes a 30° angle with the vertical, the vertical distance from the swing to the lowest level in the circular arc is ( ) cos 30 1 cos30 yLL L ° = ° and the potential energy is given by ( ) ( )( )( ) 1 cos 30 40 N 2.0 m 1 cos 30 g PE mgy mgL − ° = = 11 J . (c) At the bottom of the circular arc, y 0 = . Hence, g PE = 0 . 5.34 At maximum height, v and =0 y ( ) = 40 m s cos60 20 m s xi x vv = . Thus, 20 m s fx y v =+ = . Choosing 0 g PE = at the level of the launch point, conservation of mechanical energy gives fi PE KE f = , and the maximum height reached is ( ) ( ) ( ) 2 40 m s 20 m s 2 29 .80 ms if f y g = 61 m .
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This note was uploaded on 01/15/2011 for the course PHYS 102 taught by Professor Vandenberg during the Fall '08 term at University of Victoria.

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T1a5s - Assignment 5 solutions(Fall term Physics 102 5.13(a...

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