Assume all measurements are in metres (since g is given as 9.8 m/s^2)
The centre of gravity of the full tank is at 1.5/2=0.75 from the bottom.
Work done is equal to the potential energy of the water at the spout (top) minus the potential of the full tank.
Volume of tank, V=(πr^2 h)
Mass of water = V*ρ
where ρ=density of water = 1000 kg/m^3
PE of water in tank, PE1
=ρV*1.5g/2
PE of water at spout, PE2
=ρV*1.5g
Therefore
work done
=PE2-PE1
=ρV(1.5-0.75)g
=0.75ρg V
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for π. If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)
r = 3
h = 1.5
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