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Question Number 181730 by neinhaltsieger369 last updated on 29/Nov/22

Commented by neinhaltsieger369 last updated on 29/Nov/22

 Help - me!

$$\:\boldsymbol{\mathrm{Help}}\:-\:\boldsymbol{\mathrm{me}}! \\ $$

Commented by mr W last updated on 29/Nov/22

work=force×distance  so to determine the work done, you   should also know the distance the   particle moved. usually you get the  distance from the position vector of  the particle. for example  r(t)=2t^3 i+3tj  then r′(t)=6ti+3j  W=∫_0 ^1 F(r(t))∙r′(t)dt       =∫_0 ^1 [−(2t^3 )^2 ×3t×6t+2t^3 ×(3t)^2 ×3]dt       =∫_0 ^1 (−72t^8 +54t^5 )dt       =[−8t^9 +9t^6 ]_0 ^1        =−8+9=1 J

$${work}={force}×{distance} \\ $$$${so}\:{to}\:{determine}\:{the}\:{work}\:{done},\:{you}\: \\ $$$${should}\:{also}\:{know}\:{the}\:{distance}\:{the}\: \\ $$$${particle}\:{moved}.\:{usually}\:{you}\:{get}\:{the} \\ $$$${distance}\:{from}\:{the}\:{position}\:{vector}\:{of} \\ $$$${the}\:{particle}.\:{for}\:{example} \\ $$$$\boldsymbol{{r}}\left(\boldsymbol{{t}}\right)=\mathrm{2}{t}^{\mathrm{3}} {i}+\mathrm{3}{tj} \\ $$$${then}\:\boldsymbol{{r}}'\left(\boldsymbol{{t}}\right)=\mathrm{6}{ti}+\mathrm{3}{j} \\ $$$${W}=\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{F}}\left(\boldsymbol{{r}}\left(\boldsymbol{{t}}\right)\right)\centerdot\boldsymbol{{r}}'\left(\boldsymbol{{t}}\right){dt} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left[−\left(\mathrm{2}{t}^{\mathrm{3}} \right)^{\mathrm{2}} ×\mathrm{3}{t}×\mathrm{6}{t}+\mathrm{2}{t}^{\mathrm{3}} ×\left(\mathrm{3}{t}\right)^{\mathrm{2}} ×\mathrm{3}\right]{dt} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{72}{t}^{\mathrm{8}} +\mathrm{54}{t}^{\mathrm{5}} \right){dt} \\ $$$$\:\:\:\:\:=\left[−\mathrm{8}{t}^{\mathrm{9}} +\mathrm{9}{t}^{\mathrm{6}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:=−\mathrm{8}+\mathrm{9}=\mathrm{1}\:{J} \\ $$

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