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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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What work is done by a force $F=(2xN)i+(3N)j$,with $x$ in meters,that moves a particle from a position $r_i=(2m)i+(3m)j$ to a position $r_f=-(4m)i-(3m)j$?

$\begin{array}{1 1}-6J\\-5J\\6J\\-15J\end{array} $

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Answer : -6J
We use the general definition of work (for a two-dimensional problem),
$W=\int\limits_{x_i}^{x_f}F_x(r)dx+\int\limits_{y_i}^{y_f} F_y(r)dy$
With $F_x=2x$ and $F_y=3$
$W=\int\limits_{2m}^{-4m}2xdx+\int\limits_{3m}^{-3m}3dy$
$\;\;\;\;=x^2\big|_{2m}^{-4m}+3x\big|_{3m}^{-3m}$
$\;\;\;\;=[(16)-(4)]J+[(-9)-(9)]J$
$\;\;\;\;=-6J$
answered Sep 1, 2014 by sreemathi.v
 

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