ch07-p066 - 66. From Eq. 7-32, we see that the area in the...

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KK W Kx xx x x f f ff −= << 33 12 4 30 () ( . ) so that the requirement 8.0 J xf K = leads to x f = 40 . m . (c) As long as the work is positive, the kinetic energy grows. The graph shows this situation to hold until x = 1.0 m. At that location, the kinetic energy is 100 1 16 J 2.0 J 18 J. x KKW =+ = + = 66. From Eq. 7-32, we see that the “area” in the graph is equivalent to the work done. We find the area in terms of rectangular [length × width] and triangular [ 1 2 base × height] areas and use the work-kinetic energy theorem appropriately. The initial point is taken to
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This note was uploaded on 03/19/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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