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

As the engine is moving up the slope, force of friction `f=muR=mu mg cos theta` acts down the plane, figure. The upward force the engine has to apply.
`F=mg sin theta +mu mg cos theta=mg(sin theta +mu cos theta ) =10^(5)xx9.8(sin 5^(@)+0.1 cos 5^(@))`
`F=10^(5)xx9.8(0.0872+0.1xx0.9962)=10^(5)xx9.8xx0.18682=1.830836xx10^(5)N`
`upsilon=100m//h=(100)/(60xx60)m//s=(1)/(36)m//s`
Power of engine, `P=Fxxupsilon=1.830836xx10^(5)xx(1)/(36)=5085.6wat t`
Work done by engine in one hour `=Pxxt=5085.6xx60xx60J`
As efficiency `=("output")/("input")=("work done")/("energy used")`
`:.` energy used `=("work done")/("efficiency")=(5085.6xx60xx60)/(4//100)`
`=4.577xx10^(8)J`
`:.` Amount of coat burnt in one hour `=(4.577xx10^(8)J)/(5000J//kg)=9.154xx10^(4)kg`