Answer:
As the engine is moving up the slope, force of
friction \[f=\mu R=\mu \,mg\,\cos \theta \] acts down the plane. Fig.
4(HT).4. The upward force the engine has to apply
\[F={{10}^{5}}\times 9.8\left( 0.0872+0.1\times 0.9962
\right)\]
\[={{10}^{5}}\times 9.8\times 0.18682=1.830836\times {{10}^{5}}N\]
\[\upsilon =100m/h=\frac{100}{60\times 60}m/s=\frac{1}{36}m/s\]
Power of engine, \[P=F\times \upsilon =1.830836\times
{{10}^{5}}\times \frac{1}{36}=5085.6watt\]
Work done by engine in one hour = P x t = 5085-6 x 60 x 60 J
As \[\text{efficiency=}\frac{output}{input}=\frac{work\,done}{energy\,used}\]
\[\therefore \] \[energy\,used=\frac{work\,done}{efficiency\,}=\frac{5085.6\times
60\times 60}{4/100}=4.577\times {{10}^{8}}J\]
\[\therefore \] Amount of coal burnt in one
hour \[=\frac{4.577\times {{10}^{8}}J}{5000\,J/kg}=9.154\times {{10}^{4}}kg\]
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