A railway engine weighing 40 metic ton is travelling along a level track at a speed of `54 km H^(-1)` What additional power is required to maintain the same speed up an incline of 1 in 49 Take `g = 9.8 m//s^(2)` and `u = 0.1` .

Explanation

Here ` m =40` metric ton `= 40 xx 10^(3) kg`
`upsilon = 54 km h^(-1) = (54 xx 1000)/(60 xx 60) = 15 ms^(-1)`
Now, `cos theta = sqrt(1 - sin^(2) theta) = sqrt(1 - (1//49)^(2)) = 1`
Power required on level track ` P_(1) = F xx upsilon = mu mg xx upsilon`
Power required up an incline `P_(2) = (mg sin theta + mu mg cos theta) upsilon`
Additional power required
`P =P_(2) P_(1) = [ mg sin theta + mu mg cos theta - mu mg] upsilon = [ mg sin theta + mu mg xx 1 - mu mg] upsilon = mg sin theta xx upsilon`
`=40 xx 10^(3) xx 9.8 xx (1)/(49) xx 15 = 120 xx 10^(3) "watt" = 120 kW`.