A circular swimming pool has a diameter of 18 m, the sides are 3 m high, and the

Question

A circular swimming pool has a diameter of 18 m, the sides are 3 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^3.) How much work (in Joules) is required to: (a) pump all of the water over the side? (b) pump all of the water out of an outlet 3 m over the side? I have volume pir^2dx = pi*324dx mass = 1000*pir^2dx = pi*324dx F=mg = 9.8*1000*pir^2dx = pi*324dx W=Fd =9.8*1000*pir^2dx = 9.8*100*1000*pi*324dx(3-x) i think the distance is (3-x) then we integrate 9.8*1000*pi*324dx(3-x) from .5 to 3 but the answer i put into webwork is wrong so somewhere along the way I am not doing it right

Explanation / Answer

work done will be equal to change in potential energy.......... now pe = mgh/2....because centre of mass of water will be at 2.5/2 m from top..................W=1000*9.8*pi*9^2*2.5*2.5/2=7789162.5J.....................................................................................in this case the centre of mass of water has been shifted from 1.25m below to 3 m above so total change in height of centre of mass=4.25m thus W= change in potential energy= M*g8 rise in centre of mass =13241576.25J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.