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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A position dependent force $\overrightarrow{F}=(7-2x+3x^2)N$ acts on a small object of mass 2kg to displace it from $x=0$ to $x=5m$.The work done in joules is

$\begin{array}{1 1}70\;J\\270\;J\\35\;J\\135\;J\end{array} $

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Answer : 135 J
Work done =$\int\limits_{x_1}^{x_2}Fdx$
$\qquad\qquad=\int\limits_0^5(7-2x+3x^2)dx$
$\qquad\qquad=7x-x^2+x^3\big]_0^5$
$\qquad\qquad=35-25+125$
$\qquad\qquad=135$ J
answered Aug 28, 2014 by sreemathi.v
edited Sep 2, 2014 by sreemathi.v
 

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