Question 120167
work is force times distance ___ using the MKS system


it is a linear relationship, so the work done in moving a "section" of water
___ is the average depth [(top+bottom)/2] multiplied by the mass of the water
___ [(area)*(bottom-top)*(density)] and the gravitational acceleration (g)


a)  w=mgh __ w=((3m*6m*2m)*1000kg/m^3)(9.8nt/kg)(1m) __ w=352.8kJ


b) the average depth is 1/2 and so is the mass of water __ 352.8*.5*.5=88.2kJ


c) the average depth is 3/2 and the mass is 1/2 __ 352.8*1.5*.5=264.6kJ


d) the water in c) has to be lifted 3 times as high (on average)


e) F(y)=(y)(y/2)(176.4kJ) __ F(y)=(88.2nt/m)y^2m^2


f) 176.4kJ=88.2y^2 __ 2m^2=y^2m^2 __ y=sqrt(2)m