The problem is: A leaky 10-kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs 0.8 kg/m. Initially the bucket contains 36 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12 m level. How much work is done? Work = Force*Distance Force = Mass*Acceleration Mass = Volume*Density So, Force = Volume*Density*Acceleration don't know how relevent some of those are... For work that does not have a constant force Work = the definate intregral of [tex]\int[/tex] f(x)dx, f(x) being the force. Acceleration = 9.8 m/s^2 Density of Water = 1000 kg/m^3 I put the bounds of my integral as being [0,12]. x is the depth of the bucket so x = 0 is the top and x = 12 is the bottom 36/12 = 3 kg/m water lost Force = ((36-3x) + 10 + 0.8x)*9.8 Take the Integral of Force from 0 to 12 Answer = 3857.28 J Does this look right?