Physics
posted by blah on .
At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force that the pole exerts back on the athlete is given by F(x)= (150 N/m)x  (180N/m^2)x^2 acting over a distance of 0.20 .

integrate F dx
work from x = 0 to x = X is
W = (150/2)X^2 + (180/3)X^3
so put in 0.2 for X