Find the force required to move a train of mass `5000` quintal up an incline of `1` in `50` with an acceleration of `2m//s^(2)` Take force of friction = 0.2 N quintal and `g = 10ms^(-2)`.

Answer

`1.101 xx 10^(6)` .

Explanation

Here, `m = 5000 "quintals" = 5 xx 10^(5)kg`
`sin theta = 1//50, a = 2m//s^(2) F = 0.2N//"quintal"`
`0.2 xx 5000 = 10^(3) N`
Total force required `f = mg sin 0 + F + ma`
`f = 5 xx 10^(5) xx 10 xx (1)/(5) + 10^(3) + 5 xx 10^(5) xx 2`
`= 10^(5) + 10^(3) + 10^(6) = 10^(6) (0.1 + 0.001 + 1)`
`f = 1.101 xx 10^(6)N` .