191_PartUniversity Physics Solution

191_PartUniversity - 6-8 6.29 Chapter 6 IDENTIFY and SET UP Use Eq(6.8 to calculate k for the spring Then Eq(6.10 with x1 = 0 can be used to

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6-8 Chapter 6 6.29. IDENTIFY and SET UP: Use Eq.(6.8) to calculate k for the spring. Then Eq.(6.10), with 1 0, x = can be used to calculate the work done to stretch or compress the spring an amount 2 . x EXECUTE: Use the information given to calculate the force constant of the spring. x Fk x = gives 160 N 3200 N/m 0.050 m x F k x == = (a) (3200 N/m)(0.015 m) 48 N x x = (3200 N/m)( 0.020 m) 64 N x x = (magnitude 64 N) (b) 22 11 (3200 N/m)(0.015 m) 0.36 J Wk x = (3200 N/m)( 0.020 m) 0.64 J x = Note that in each case the work done is positive. EVALUATE: The force is not constant during the displacement so Eq.(6.2) cannot be used. A force in the x + direction is required to stretch the spring and a force in the opposite direction to compress it. The force x F is in the same direction as the displacement, so positive work is done in both cases. 6.30. IDENTIFY: The magnitude of the work can be found by finding the area under the graph. SET UP: The area under each triangle is 1/2 base height × . 0 x F > , so the work done is positive when x increases during the displacement. EXECUTE: (a) 1/2 (8 m)(10 N) 40 J = . (b) 1/2 (4 m)(10 N) 20 J = . (c) 1/2 (12 m)(10 N) 60 J = . EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c). 6.31. IDENTIFY: Use the work-energy theorem and the results of Problem 6.30. SET UP: For 0 x = to 8.0 m x = , tot 40 J W = . For 0 x = to 12.0 m x = , tot 60 J W = . EXECUTE: (a) (2)(40 J) 2.83 m/s 10 kg v (b) (2)(60 J) 3.46 m/s 10 kg v . EVALUATE: F " is always in the -direction. x + For this motion F " does positive work and the speed continually increases during the motion. 6.32. IDENTIFY: The force has only an x -component and the motion is along the x -direction, so 2 1 x x x WF d x = . SET UP: 1 0 x = and 2 6.9 m x = . EXECUTE: The work you do with your changing force is 2 1 2 ( ) ( 20.0 N) (3.0 N/m) ( 20.0 N) | (3.0 N/m)( /2) | xx x x x x x x x d x d x x d x x x = ∫∫ 138 N m 71.4 N m 209 J W =− . EVALUATE: The work is negative because the cow continues to move forward (in the -direction x + ) as you vainly attempt to push her backward. 6.33. IDENTIFY: Apply Eq.(6.6) to the box. SET UP: Let point 1 be just before the box reaches the end of the spring and let point 2 be where the spring has maximum compression and the box has momentarily come to rest. EXECUTE: tot 2 1 WK K 2 1 10 2 , K mv = 2 0 K = Work is done by the spring force. 2 1 tot 2 2 , x where 2 x is the amount the spring is compressed. 20 kx mv −= and / (3.0 m/s) (6.0 kg)/(7500 N/m) 8.5 cm xvm k = EVALUATE: The compression of the spring increases when either 0 v or m increases and decreases when k increases (stiffer spring). 6.34. IDENTIFY: The force applied to the springs is x x = . The work done on a spring to move its end from 1 x to 2 x is 21 x k x . Use the information that is given to calculate k . SET UP: When the springs are compressed 0.200 m from their uncompressed length, 1 0 x = and 2 0.200 m x . When the platform is moved 0.200 m farther, 2 x becomes 0.400 m .
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Work and Kinetic Energy 6-9 EVALUATE: (a) 22 2 21 ( 8 0 . 0 J ) 4000 N/m (0.200 m) 0 W k xx == = −− . (4000 N/m)( 0.200 m) 800 N x Fk x = . The magnitude of force that is required is 800 N.
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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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191_PartUniversity - 6-8 6.29 Chapter 6 IDENTIFY and SET UP Use Eq(6.8 to calculate k for the spring Then Eq(6.10 with x1 = 0 can be used to

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