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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A train of mass $2.0 \times 10^5$ kg has constant speed of $20 m/s$ up a hill inclined at $\theta, (\sin \theta=\large\frac{1}{50})$ to the horizontal when the engine is working at $8.0 \times 10 ^5 $ watt. Find the resistance to the motion of the train $(g=9.8 m/s^2)$

\[(a)\;R=400 \; N\quad (b)\;R=800 \; N\quad (c)\; R=1500\;N \quad (d)\;R=200\;N\]

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Power= F.v
$F=\large\frac{p}{v}=\frac{8 \times 10 ^5}{20}$$=4.0 \times 10^4 N$
At constant speed forces acting on the train are in equillibrium.
Resolving the forces parallel to the hide
$F=R+mg \sin \theta$
$4.0 \times 10^4=R+(2.0 \times 10^5) \times 5.8 \times \large\frac{1}{50}$
$R=800\; N$
Hence b is the correct answer.


answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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