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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A particle of mass 2 kg moves on a horizontal plane under the action of force $F=(2 \hat i+2 \hat j)N$. The body is displaced from (0,0) to (1m,1m) If the initial speed of the particle is 2m/s find its final speed

\[(a)\;2 \sqrt 2\;m/s \quad (b)\;4 \;m/s \quad (c)\;4 \sqrt 2 m/s \quad (d)\;5\;m/s\]

Can you answer this question?
 
 

1 Answer

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From work energy theorm
$W=\Delta KE$
$W=\bar F. \bar S$
$\quad=(2 \hat i+2 \hat j).(\hat i+\hat j)$
$\quad=4\;J$
$4=\large\frac{1}{2}$$mv^2-\large\frac{1}{2}$$m(2)^2$
$8=2 \times v^2-8$
$v=2 \sqrt 2 m/s$
Hence a is the correct answer.

 

answered Jul 22, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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