F10P1AChap7Sol

F10P1AChap7Sol - Potential Energy SOLUTIONS TO PROBLEMS...

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1 Potential Energy SOLUTIONS TO PROBLEMS Section 7.1 Potential Energy of a System P7.2 (a) We take the zero configuration of system potential energy with the child at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus, U g = mgy = 400 N ( ) 2.00 m ( ) = 800 J . (b) From the sketch, we see that at an angle of 30.0° the child is at a vertical height of 2.00 m ( ) 1 ! cos30.0 ° ( ) above the lowest point of the arc. Thus, FIG. P7.2 U g = mgy = 400 N ( ) 2.00 m ( ) 1 ! cos30.0 ° ( ) = 107 J . (c) The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at this location.
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2 Potential Energy Section 7.2 The Isolated System P7.5 (a) U i + K i = U f + K f : mgh + 0 = mg 2 R ( ) + 1 2 mv 2 g 3.50 R ( ) = 2 g R ( ) + 1 2 v 2 v = 3.00 gR (b) F ! = m v 2 R : n + mg = m v 2 R n = m v 2 R ! g " # $ % & ' = m 3.00 gR R ! g " # $ % & ' = 2.00 mg n = 2.00 5.00 ( 10 ! 3 kg ( ) 2 ( ) = 0.098 0 N downward FIG. P7.5 P7.7 From leaving ground to the highest point, K i + U i = K f + U f 1 2 m ( ) 2 + 0 = 0 + m 9.80 m s 2 ( ) y The mass makes no difference: ! y = ( ) 2 2 ( ) 9.80 m s 2 ( ) = 1.84 m P7.9 Using conservation of energy for the system of the Earth and the two objects (a) 5.00 kg ( ) g 4.00 m ( ) = 3.00 kg ( ) g 4.00 m ( ) + 1 2 5.00 + 3.00 ( ) v 2 v = 19.6 = 4.43 m s (b) Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its free fall.
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This note was uploaded on 11/27/2011 for the course PHYS 1A taught by Professor Onuchic during the Fall '07 term at UCSD.

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F10P1AChap7Sol - Potential Energy SOLUTIONS TO PROBLEMS...

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