Question

(2,4 A water tank is formed by revolving this graph about It is 4 feet wide and 4 ft high, as shown in the picture. ater is to be pumped to a level one foot above the top, work will be required? Let w be the weight density, in lb per ft", so that your answer will be an expression number

Answer 1

y=x² B The water took formed by a parabolloid. defined by revoking the graph z-me and 7 = 4 2-4 O<7 <2 OKO <2IT. potential energy E = righ m mass acceleration 9.81 m/s2 9-> goou. the height MPV WV w density of water af. initial potential energes- Ei Spargez Eisspg z dv o dodraz 2 DIT 7 0 4 = 211 ss ndodz. Ei = w.Rod Povoar sazda awang Sindr (44) wangstjerde - J**.cdr] og [ 4? [ en ] The 2 8 0 IT W 128 0 wq 5 mass of wather in lank, m = P San => ME jer jes" ndo do dz w . 0 - W 2T je rodo s dz 12

ma 2 IT W Twf rodo (4-0) -20[4.1vdv - Jonido] = 25$ 461-17] 8Tw final potential emerges, Ef = mgh h = height of water teach - 5 ft. 40 go. stwig igis 128 T 5 LWA required work done difference Ice between energies. >> W= 4 £ = £f - £i 40 tr qwa = (40 - 128 ) tog 72 7 wg Ab-ft W = put. 4 - 82 ft 182, we get, W = 1447.640 lb-ft Answey work done