Find the force required to move a train of mass `10^(5)` kg up an incline of 1 in 50 with an acceleration of `2 ms^(-2)`. Coefficient of friction between the train and rails is 0.005. Take `g = 10^(2)`.

Find the force required to move a train of mass `10^(5)` kg up an incl

1.	Find the force required to move a train of mass `10^(5)` kg up an incline of 1 in 50 with an acceleration of `2 ms^(-2)`. Coefficient of friction between the train and rails is 0.005. Take `g = 10^(2)`.
Answer» ` F = ? , m = 10^(5) kg , sin theta (1)/(50) , ` ` a = 2 m // s^(2) , mu = 0.005 , g = 10 ms^(-2) ` Now `cos theta = sqrt ( 1 - sin^(2) theta) = sqrt (1 - ((1)/(50))^(2))= sqrt((2499)/(2500))=1` ` F = mg (sin theta + mu cos theta ) + ma ` ` = 10^(5) xx 10 ((1)/(50) + 0.005 xx 1 ) + 10^(5) xx 2 ` ` 10^(6) xx 0 .025 + 2 xx 10^(5) = 2.25 xx 10^(5) N ` .

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Find the force required to move a train of mass `10^(5)` kg up an incline of 1 in 50 with an acceleration of `2 ms^(-2)`. Coefficient of friction between the train and rails is 0.005. Take `g = 10^(2)`.

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