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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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An objecte of mass 5 kg falls from rest through a vertical distance of 20 m and reaches velocity 10 m/s. How much work is done by push of air on the object $(g=10 m/s^2)$

\[(a)\;120\;J\quad (b)\;-750\;J \quad (c)\; -450\;J \quad (d)\;250\; J\]

Can you answer this question?
 
 

1 Answer

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From work energy theorem work done by weight + work done by air $= \Delta KE$
$W_{\large air}=\Delta KE-W_{(\large weight)}$
$ \qquad=\large\frac{1}{2}$$mv^2-mgh$
$\qquad=\large\frac{1}{2} $$ \times 5 \times (10)^2-5 \times 10 \times 20$
$\qquad=-750\;J$
Hence b is the correct answer. 
answered Aug 7, 2013 by meena.p
edited Jun 11, 2014 by lmohan717
 

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