A 100000 kg engine is moving up a slope of gradient 5 deg at a speed of 100 mt / hour. The co-efficient of friction between the engine & rails is 0.1. If the engine has an efficient of 4 % for converting heat into work. Find the amount of coal the engine has to burn up in one hour. ( Burning of 1 kg coal yields 50000 joule )

Dear Student,

m=105 kgv=100 m/hθ=5°g=9.8 m/s2μ=0.1Total force applied by the engine, F=mg sinθ+μRF=mg sinθ+μ mg cosθF=mgsinθ+μ cosθTotal power, P=mgsinθ+μ cosθv



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