A train of mass `2xx10^(5)` kg has a constant speed of 20 `ms^(-1)` up a hill inclined at `theta=sin^(-1)((1)/(50))` to the horizontal when the engine is working at `8xx10^(5)` W. Find the resistance to motion of the trian. (Take ,g=9.8 `ms^(-2)`).

Explanation

Since, P=Fv
`F=(P)/(v)=(8xx10^(5))/(20)=4xx10^(4) N`
At constant speed, the forces acting on the train are in equilibrium. Resolving the forces parallel to the hill

`F=R+(2.0xx10^(5))gxx(1)/(50)`
`4xx10^(4)`=R+39200
or `" " R=800 N`
Therefore, the resistance is 800 N.