Correct Answer - `(i) 10^(4) W (ii) 2960 N` .
Here, `m = 1200 kg, upsilon = 10m//s, T =1000 N`
`P = ? sin theta = (1)/(6), T = ?`
One the level,force applied by the truck = force of friction overcome
`:. F = T = 1000N`
power expended on the trailor
`P = f xx upsilon = 1000 xx 10 = 10^(4)W`
When the truck ascends a road it has to overcome friction and component `mg sin theta` of the weight of trailor.
`F = mg sin theta + f`
`1200 xx 9.8 xx (1)/(6) + 1000 = 1960 + 1000`
`=2960 N` .