Question

A particle moves along the x-axis from x=0 to x=5 m under the influence of a force given by $\sf {F=7−2x+3x^{2}}$ The work done in the process is ?

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Answer 1

❥Heyo Meoww✌️!

➻If you're good at integration, you're gonna crack this in a giffy ;]

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$\rm \red {\maltese{\underline{\underline{Provided:}}}}$

➻So we've been given the limits from 0 to 5.

$\sf {⇨Initial\: limit=0}$

$\sf {⇨Final\: limit=5}$

$\sf {⇨Force(F)=7−2x+3x^{2}}$

$\rm \blue {\maltese{\underline{\underline{To\: Find:}}}}$

➻The work done in the process of displacement is?

$\rm \pink {\maltese{\underline{\underline{We\: Know:}}}}$

$\sf \implies {\boxed{\underline{W=\displaystyle \int F\: dx}}}$

$\sf \to {\displaystyle \int\limits_{0}^{5} (7−2x+3x^{2})\: dx}$

$\sf \to {\displaystyle \int\limits_{0}^{5} 7\: dx-\displaystyle \int\limits_{0}^{5}2x\: dx+ \displaystyle \int\limits_{0}^{5}3x^{2}\: dx}$

$\large \sf \to {[7x]_{0}^{5}-[\frac{2x^{2}}{2}]_{0}^{5}+[\frac{3x^{3}}{3}]_{0}^{5}}$

$\large \sf \to {[7x]_{0}^{5}-[\frac{\cancel{2}x^{2}}{\cancel{2}}]_{0}^{5}+[\frac{\cancel{3}x^{3}}{\cancel{3}}]_{0}^{5}}$

$\sf \to {7(5-0)-(5)^{2}-0+(5)^{3}-0}$

$\sf \to {7(5)-25+125}$

$\sf \to {35-25+125}$

$\sf \to {10+125}$

$\sf \implies \green {\boxed{\boxed{135J}}}$

$\rm \purple {\maltese{\underline{\underline{Henceforth,}}}}$

➻The work done in the process is '135 joules' respectively.

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ᵜpew pewᵜ !

⍟Take care n study well :)<3