Calculating Work and Power from Ocean Water Pumping

[Ogakor]

Homework Statement

Water is pumped from the ocean to a large reservoir on land. A total lift of 140m. At a rate of 60 cubic meters per hour and eject it with a speed of 65 m/s. If one cubic meter of sea water has a mass of 1,025 kg, find:
a) The work done in lifting the water
b) the work done in ejecting it
c) the total work done
d) the power developed
e) if the pump is 90% efficient, what horse power engine is needed

given:
d = 140m
r (rate) = 60 cubic meters per hour
v = 65 m/s
m = 1,025 kg

for sure, A (work done in lifting the water) can be solved with W (work) = Fd.
Since F is not given, W = (ma)d
since a is not given, a = -v₁² / 2d.

my dad said B (work done in ejecting it) can be solved with KE = 1/2mv². But I am not sure.

Total work can be solved by adding A and B.

I don't know how to solve D and E.

Are my fomulas for A, B and C correct? What formulas should I use for D and E?
Please and Thanx! :)

 

[rock.freak667]

given:
d = 140m
r (rate) = 60 cubic meters per hour
v = 65 m/s
m = 1,025 kg

for sure, A (work done in lifting the water) can be solved with W (work) = Fd.
Since F is not given, W = (ma)d
since a is not given, a = -v₁² / 2d.

my dad said B (work done in ejecting it) can be solved with KE = 1/2mv². But I am not sure.

It is much easier to use energy indeed.

you were given ρ=1025 kg/m³, not 'm'.

If you have the volume flow rate of 60 m³/hr, what do you get if you multiply that by the density (convert the volume flow rate to m³/s and then multiply)?

If the energy needed to lift the water a distance 'h' is E=mgh and we differentiate w.r.t. t, we will get dE/dt = d/dt(mgh) = gh*dm/dt. We know that power P=dE/dt, so P=gh*dm/dt.