Answer:
Here,
\[m=40\,\]metric ton = \[40\times {{10}^{3}}kg\]
\[\upsilon
=54km\,{{h}^{-1}}=\frac{54\times 1000}{60\times 60}=15\,m\,{{s}^{-1}}\]
\[\sin \theta =\frac{1}{49}\]; \[\mu =0.1\]; \[g=10m{{s}^{-2}}\]
Now, \[\cos \theta
=\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{\left( 1/49 \right)}^{2}}}\approx 1\]
Power required on level track, \[{{P}_{1}}=F\times
\upsilon =\mu \,mg\times \upsilon \]
Power required up an incline, \[{{P}_{2}}=\left(
mg\,\sin \theta +\mu \,mg\,\cos \theta \right)\upsilon \]
Additional power required
\[P={{P}_{2}}-{{P}_{1}}=\left[
mg\,\sin \theta +\mu \,mg\,\cos \theta -\mu mg \right]\upsilon \]
\[=\left[ mg\,\sin \theta +\mu mg\times 1-\mu mg
\right]\upsilon =mg\,\sin \theta \times \upsilon \]
\[P=40\times
{{10}^{3}}\times 9.8\times \frac{1}{49}\times 15=120\times
{{10}^{3}}watt=120kW\]
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