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191_PartUniversity Physics Solution

# 191_PartUniversity Physics Solution - 6-8 6.29 Chapter 6...

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6-8 Chapter 6 6.29. I DENTIFY and S ET U P : 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 E XECUTE : Use the information given to calculate the force constant of the spring. x F kx = gives 160 N 3200 N/m 0.050 m x F k x = = = (a) (3200 N/m)(0.015 m) 48 N x F kx = = = (3200 N/m)( 0.020 m) 64 N x F kx = = = − (magnitude 64 N) (b) 2 2 1 1 2 2 (3200 N/m)(0.015 m) 0.36 J W kx = = = 2 2 1 1 2 2 (3200 N/m)( 0.020 m) 0.64 J W kx = = = Note that in each case the work done is positive. E VALUATE : 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. I DENTIFY : The magnitude of the work can be found by finding the area under the graph. S ET U P : 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. E XECUTE : (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 = . E VALUATE : The sum of the answers to parts (a) and (b) equals the answer to part (c). 6.31. I DENTIFY : Use the work-energy theorem and the results of Problem 6.30. S ET U P : For 0 x = to 8.0 m x = , tot 40 J W = . For 0 x = to 12.0 m x = , tot 60 J W = . E XECUTE : (a) (2)(40 J) 2.83 m/s 10 kg v = = (b) (2)(60 J) 3.46 m/s 10 kg v = = . E VALUATE : F " is always in the -direction. x + For this motion F " does positive work and the speed continually increases during the motion. 6.32. I DENTIFY : The force has only an x -component and the motion is along the x -direction, so 2 1 x x x W F dx = . S ET U P : 1 0 x = and 2 6.9 m x = . E XECUTE : The work you do with your changing force is 2 2 2 2 2 1 1 1 1 1 2 ( ) ( 20.0 N) (3.0 N/m) ( 20.0 N) | (3.0 N/m)( /2) | x x x x x x x x x x W F x dx dx xdx x x = = = − 138 N m 71.4 N m 209 J W = − = − . E VALUATE : 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. I DENTIFY : Apply Eq.(6.6) to the box. S ET U P : 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. E XECUTE : tot 2 1 W K K = 2 1 1 0 2 , K mv = 2 0 K = Work is done by the spring force. 2 1 tot 2 2 , W kx = − where 2 x is the amount the spring is compressed. 2 2 1 1 2 0 2 2 kx mv = − and 2 0 / (3.0 m/s) (6.0 kg)/(7500 N/m) 8.5 cm x v m k = = = E VALUATE : The compression of the spring increases when either 0 v or m increases and decreases when k increases (stiffer spring). 6.34. I DENTIFY : The force applied to the springs is x F kx = . The work done on a spring to move its end from 1 x to 2 x is 2 2 1 1 2 1 2 2 W kx kx = . Use the information that is given to calculate k . S ET U P : When the springs are compressed 0.200 m from their uncompressed length, 1 0 x = and 2 0.200 m x = − .

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