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A railway engine weighing 40 metric ton ...

A railway engine weighing 40 metric 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 `m u = 0.1`.

Text Solution

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Here , `m = 40` metric ton
` = 40 xx 10^(3) kg`
`upsilon 54 km//h = (54 xx 1000) /( 60 xx 60 ) m //s = 15 m//s`,
`g = 9.8 m // s^(2)`
`mu = 0. 1 , sin theta = 1//49 , (P_(2) - P_(2) ) = ? `
On a level track ,
`P_(1) = mu mg xx upsilon = 0.1 xx 4 xx 10^(4) xx 9.8 xx 15`
`5.88 xx 10^(5) "watt"`
Up the incline , ` P_(2) = mg (sin theta + mu cos theta) xx upsilon`
` 4 xx 10^(4) xx 9.8 ((1)/(49) + 0.1 xx 1 ) xx 15 `
` 7 .08 xx 10^(5) W `
Additional power required
` = P_(2) - P_(1) = (7 .08 - 5 .88 ) 10^(5) W `
` = 1.20 xx 10^(5) W = 120 k W ` .
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