Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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

$$\mathrm{Time}\:\mathrm{taken}\:\mathrm{1}\:\mathrm{second}. \\ $$$$\mathrm{Potential}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{P}} \:: \\ $$$$\mathrm{E}_{\mathrm{P}} \:=\:{mgh}\:=\:\mathrm{50}×\mathrm{9},\mathrm{81}×\mathrm{40}\:=\:\mathrm{19620}\:\mathrm{J} \\ $$$$\mathrm{Kinetic}\:\mathrm{energy}\:\mathrm{E}_{\mathrm{K}} \:: \\ $$$$\mathrm{E}_{\mathrm{K}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{50}×\mathrm{8}^{\mathrm{2}} \:=\:\mathrm{1600}\:\mathrm{J} \\ $$$$\mathrm{Now},\:\mathrm{the}\:\mathrm{work}\:\mathrm{done}\:\mathrm{per}\:\mathrm{second}\:\mathrm{is}\:: \\ $$$${P}_{{u}} \:=\:\frac{\mathrm{19620J}+\mathrm{1600J}}{\mathrm{1}{s}}\:=\:\mathrm{21220W} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com