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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A constant force $F=(2\hat{i}+3\hat{j})$ newton displaces a body from position $r_1=(4\hat{i}-5\hat{j})$ metre to $r_2=(\hat{i}+3\hat{j})$ metre.Find the work done by the force.

$\begin{array}{1 1}18J\\15J\\16J\\20J\end{array} $

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Answer : 18J
Displacement $S=r_2-r_1$
$\qquad\qquad\quad\;\;=(\hat{i}+3\hat{j})-(4\hat{i}-5\hat{j})$
$\qquad\qquad\quad\;\;=-3\hat{i}+8\hat{j}$
$W=F.S$
$\;\;\;\;=(2\hat{i}+3\hat{j}).(-3\hat{i}+8\hat{j})$
$\hat i.\hat i=\hat j.\hat j=1$ and $\hat i.\hat j=0$
$\;\;\;\;=-6+24$
$\;\;\;\;\;=18$J
answered Aug 28, 2014 by sreemathi.v
 

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