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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Due to the application of a force body of mass 1.5 kg is moving up vertically with an acceleration of $1.2 m/s^2$ for $10 \;s$, starting from rest, What is power of the agent doing work?

\[(a)\;99\;W \quad (b)\;990\;W \quad (c)\;9\;W \quad (d)\;1.2\;W\]
Can you answer this question?
 
 

1 Answer

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Force 'F' acting upwards produces acceleration 'a' on it
$F-mg=ma$
$F=m(g+a)$
$\quad=1.5(9.8 +1.2)$
$\quad=16.5 \;N$
Displacement in 10 s
$s=ut+\large\frac{1}{2}$$ at^2$
$\quad=0+\large\frac{1}{2}$$ \times 1.2 \times 10^2$
$\quad=60\;m$
Power $=\large\frac{F.S}{t}$
$\quad=\large\frac{16.5 \times 60}{10}$
$\quad=99\;W$
Hence a is the correct answer.

 

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

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