Let's consider the circular cross-section through the tank at depth h and find its radius using similar triangles:
r2=10−h10,
r=102(10−h)=510−h.
The aea of the cross-section: A=πr2=25π(10−h)2.
The volume of the section of water at depth h:
V=πr2dh=25π(10−h)2dh.
The density of the water: σ=1000kg/m3, so the mass of water in the section is
m=σV=100025π(10−h)2dh=40π(10−h)2dh.
To raise this mass of water we need to overcome the force of gravity F=mg , g=9.8m/s2.
F=9.8⋅40π(10−h)2dh=392π(10−h)2dh.
We must lift this water a distance h and require a work
W=392π(10−h)2hdh.
The total work:
W=∫010392π(10−h)2hdh=392π∫010(10−h)2hdh=
=392π∫010(100h−20h2+h3)dh=392π(1002h2−203h3+4h4)∣010=
=392π(5000−320,000+410,000)=392π32500=3980,000π≈1,026kJ.
Answer: 1,026 kJ.
Comments