Question

An engine of one metric ton is going up an inclined plane of slope 1 in 2 at the rate of `36km h^(-1)` If the coefficient of friction is `1//sqrt3` calculate the power of the engine in k W .

Answer 1

Correct Answer - `98kW` .
Here, `m = 1` metric ton `= 10^(3) kg`
`sin theta = 1//2, upsilon = 36 km//h = (36 xx1000)/(60 xx 60) m//s = 10m//s`
`mu = (1)/sqrt 3 P = ?`
`cos theta = sqrt( 1 - sin^(2) theta) = sqrt(1-(1)/(4))=sqrt3/(2)`
As is known force required
`F = mg sin theta + f = mg sin theta + mu R`
`= mg sin theta + mu mg cos theta`
`= mg (sin theta + mu cos theta)`
`= 10^(3) xx 9.8 ((1)/(2) + (1)/sqrt3 sqrt3/(2))`
` F = 9.8 xx 10^(3) xx 10 "watt"`
`P = 98 kW` .