KKWKxxxxx ffff−=−−<<3312430()(.)so that the requirement 8.0 JxfK=leads to xf=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 100116 J 2.0 J 18 J.xKKW=+=+=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 [12base×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.