Calculating Integrals of Force for Work Done

[Echo88]

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 relevant some of those are...

For work that does not have a constant force
Work = the definate intregral of \int 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?

 

[D H]

Everything but the rope itself looks good.