An engine of mass `6.5` metric ton is going up an incline of 5 in 13 at the rate of `9km//h` Calculate the power of the engine if `mu = 1//2` and `g = 9.8m//s^(2)`.

Answer

`73.5 kW`

Explanation

Here, `m = 6.5` metric ton = 6500 kg
`sin theta = 5//13`
`upsilon = 9km//h = (9 xx 1000)/(60 xx 60) =2.5m//s P = ? mu = (1)/(12)`
When `sin theta = (5)/(13), cos theta = sqrt(1 - (5//13)^(2)) = 12//13`
`P = mg (sin theta + mu cos theta) xx upsilon`
`6500 xx 9.8((5)/(13) + (1)/(12) +(12)/(13)) xx 2.5`
`= 6500 xx 9.8 xx (6)/(13) xx 2.5 = 73500 W = 73.5 kW` .