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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A body of 0.02 kg falling from height 5 m into a pile of sand.The body penetrates the sand a distance of 5cm before stopping what force has the sand exerted on the body?

\[(a)\;1.96\;N \quad (b)\;-19.6 \;N \quad (c)\;-0.196 \;N \quad (d)\;0.0196\;N\]
Can you answer this question?
 
 

1 Answer

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v-velocity with which the body reaches the sand
$v^2=u^2+2as$
$\quad=2 \times 9.8 \times 5$
$\quad=98$
It penetrates send 0.05 m and comes to stop
$a=\large\frac{-98}{2 \times 0.05}$
$\quad= -980 m/s^2$
retarding force$= m \times a$
$\qquad=0.02 \times -980$
$\qquad=-19.6 N$
Hence b is the correct answer. 

 

answered Jul 12, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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