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Question-152682




Question Number 152682 by alcohol last updated on 31/Aug/21
Answered by Olaf_Thorendsen last updated on 31/Aug/21
Time taken 1 second.  Potential energy E_P  :  E_P  = mgh = 50×9,81×40 = 19620 J  Kinetic energy E_K  :  E_K  = (1/2)mv^2  = (1/2)×50×8^2  = 1600 J  Now, the work done per second is :  P_u  = ((19620J+1600J)/(1s)) = 21220W
$$\mathrm{Time}\:\mathrm{taken}\:\mathrm{1}\:\mathrm{second}. \\ $$$$\mathrm{Potential}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{P}} \:: \\ $$$$\mathrm{E}_{\mathrm{P}} \:=\:{mgh}\:=\:\mathrm{50}×\mathrm{9},\mathrm{81}×\mathrm{40}\:=\:\mathrm{19620}\:\mathrm{J} \\ $$$$\mathrm{Kinetic}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{K}} \:: \\ $$$$\mathrm{E}_{\mathrm{K}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{50}×\mathrm{8}^{\mathrm{2}} \:=\:\mathrm{1600}\:\mathrm{J} \\ $$$$\mathrm{Now},\:\mathrm{the}\:\mathrm{work}\:\mathrm{done}\:\mathrm{per}\:\mathrm{second}\:\mathrm{is}\:: \\ $$$${P}_{{u}} \:=\:\frac{\mathrm{19620J}+\mathrm{1600J}}{\mathrm{1}{s}}\:=\:\mathrm{21220W} \\ $$

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