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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A man pushes a truck initially at rest and weighting $2.5$ tonne along a horizontal rail with a steady force of $200 N$. The resistance to the motion amounts to $30 N$ per tonne. The velocity of the truck at the end of 30 seconds is

\[(a)\;1.5 m/s \quad (b)\;0.15 m/s \quad (c)\;15 m/s \quad(d)\;150 m/s \]
Can you answer this question?

1 Answer

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Total resistance offered $=\large\frac{30 N}{tonne} $$\times 2.5 \;tonne$
$\qquad=75 N$
Resultant force $= 200-75=125 N$
mass $=2.5 tonne=2500 kg$
$a=\large\frac{F}{m}=\frac{125}{2500}=\frac{1}{20} $$m/s^2$
$v=u+at =0+\large\frac{1}{20} $$\times 30$
$\quad= 1.5 m/s$
Hence a is the correct answer


answered Jul 4, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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