A cubical vessel of height 1m is full of water. The minimum work done in taking water out from vessel will be? *please explain clearly*
Answers
Answered by
492
Work = Change in Potential energy
= Final Potential energy - Initial Potential energy
= mgL - mgL/2
= mgL/2
= (ALd gL)/2
= (1^2 * 1 * 10^3 * 10 * 1)/2
= 5 kJ
= Final Potential energy - Initial Potential energy
= mgL - mgL/2
= mgL/2
= (ALd gL)/2
= (1^2 * 1 * 10^3 * 10 * 1)/2
= 5 kJ
Anonymous:
how does L/2 come?
Height of centre of mass.
No bro
u cannot do like that
the graph of the work done is a continuous function
Thanks
Centre of mass SHOULD BE considered. In your method you should take limits from h/2 to h. Because initially mass of water is considered to be at centre of cube and not at “0”
Answered by
298
volume of cubic vessel = 1m³ density of water = 1000 kg/m³ So, m = 1000 kg ∂U = mg∂x ∫∂U = mg∫∂x limits on the right hand side are from h/2 to h U = mgh - mgh/2 U = mgh/2 U = 1000• 10•1/2 = 5000 J
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