A 100 metric ton 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 50000J.

Explanation

Here, `m=100`metric ton
`=100xx1000kg=10^(5)kg`
`theta=5^(@), v=100m//h, mu=0.1eta=4%`
Mass of coal, `m=?t=1h=60xx60s`
Energy yielded `=50000J//kg` of coal.
As is clear from figure, force applied by the engine,
`F=mg sintheta+f`
`=mg sintheta +mu mg cos theta`
`=mg(sintheta+mu cos theta)`
`=10^(5)xx9.8(sin5^(@)+0.1cos 5^(@))`
`F=10^(5)xx9.8(0.0872+0.1xx0.9962)`
`=9.8xx10^(5)xx0.18682`
Power of the engine, `P=Fxxv`
`9.8xx10^(5)xx0.18683xx(100)/(60xx60)`
`=4808W`
As efficiency `=("output")/("Input")=("Work done")/("energy used")`
`:.` energy used `=("work done")/("efficiency")=(Pxxt)/("efficiency")`
`=(4808xx60xx60J)/(4//100)=4.33xx10^(8)J`
As 50000 J energy comes from burning of 1 kg of coal.
`:.` Mass of coal to be burnt
`=(4.33xx10^(8))/(50000)=8.66xx10^(3)kg`