question_answer

A 100000 kg engine is moving up a slope of gradient 5° at a speed of 100 metre/hour. The coefficient of friction between the engine and the rails is 0.1. If engine has an efficiency of 4% for converting heat into work, find the amount of coal the engine has to burn in one hour. Burning of 1 kg coal yields 50000 J.

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 \[=\frac{4.577\times {{10}^{8}}J}{5000\,J/kg}=9.154\times {{10}^{4}}kg\]