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(D) The work done by gravity is equal to the decrease in the potential energy of the system,$W = U_i - U_f$.Initial potential energy $U_i = U_1 + U_2

Vedclass · Accepted Answer

$8 	imes 10^4 \ J$

Vedclass · Answer

(D) The work done by gravity is equal to the decrease in the potential energy of the system,$W = U_i - U_f$.Initial potential energy $U_i = U_1 + U_2 = (m_1 g h_{cm1}) + (m_2 g h_{cm2})$.Given $A = 2 \ m^2$,$ho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$.$U_i = (ho A 	imes 10) g 	imes (10/2) + (ho A 	imes 6) g 	imes (6/2) = ho Ag (50 + 18) = 68 ho Ag$.When connected,the water levels equalize to $h = (10 + 6) / 2 = 8 \ m$ in both vessels.Final potential energy $U_f = (ho A 	imes 8) g 	imes (8/2) + (ho A 	imes 8) g 	imes (8/2) = ho Ag (32 + 32) = 64 ho Ag$.Work done $W = U_i - U_f = 68 ho Ag - 64 ho Ag = 4 ho Ag$.$W = 4 	imes 10^3 	imes 2 	imes 10 = 8 	imes 10^4 \ J$.

Two cylindrical vessels of equal cross-sectional area of $2 \ m^2$ contain water up to heights of $10 \ m$ and $6 \ m$,respectively. If the vessels are connected at their bottom,then the work done by the force of gravity is (Density of water is $10^3 \ kg/m^3$ and $g = 10 \ m/s^2$):

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