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Question Number 22990 by Tinkutara last updated on 24/Oct/17

A train weighing 100 metric ton is  running on a level track with a uniform  speed of 72 km h^(−1) . If the frictional  resistance amounts to 0.5 kg per metric  ton, find the power of the engine.

$$\mathrm{A}\:\mathrm{train}\:\mathrm{weighing}\:\mathrm{100}\:\mathrm{metric}\:\mathrm{ton}\:\mathrm{is} \\ $$$$\mathrm{running}\:\mathrm{on}\:\mathrm{a}\:\mathrm{level}\:\mathrm{track}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform} \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{72}\:\mathrm{km}\:\mathrm{h}^{−\mathrm{1}} .\:\mathrm{If}\:\mathrm{the}\:\mathrm{frictional} \\ $$$$\mathrm{resistance}\:\mathrm{amounts}\:\mathrm{to}\:\mathrm{0}.\mathrm{5}\:\mathrm{kg}\:\mathrm{per}\:\mathrm{metric} \\ $$$$\mathrm{ton},\:\mathrm{find}\:\mathrm{the}\:\mathrm{power}\:\mathrm{of}\:\mathrm{the}\:\mathrm{engine}. \\ $$

Answered by ajfour last updated on 24/Oct/17

let metric ton=M  resistance of 0.5kg/M means  resistance equal to (weight of  0.5kg)/M=(0.5kg×9.8m/s^2 )/M  =4.9N/M  net resistance = ((4.9N)/M)×100M  =490N      P=Fv = 490N×((5/(18))×((72m)/s))                   =9800W =9.8kW .

$${let}\:{metric}\:{ton}={M} \\ $$$${resistance}\:{of}\:\mathrm{0}.\mathrm{5}{kg}/{M}\:{means} \\ $$$${resistance}\:{equal}\:{to}\:\left({weight}\:{of}\right. \\ $$$$\left.\mathrm{0}.\mathrm{5}{kg}\right)/{M}=\left(\mathrm{0}.\mathrm{5}{kg}×\mathrm{9}.\mathrm{8}{m}/{s}^{\mathrm{2}} \right)/{M} \\ $$$$=\mathrm{4}.\mathrm{9}{N}/{M} \\ $$$${net}\:{resistance}\:=\:\frac{\mathrm{4}.\mathrm{9}{N}}{{M}}×\mathrm{100}{M} \\ $$$$=\mathrm{490}{N} \\ $$$$\:\:\:\:{P}={Fv}\:=\:\mathrm{490}{N}×\left(\frac{\mathrm{5}}{\mathrm{18}}×\frac{\mathrm{72}{m}}{{s}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9800}{W}\:=\mathrm{9}.\mathrm{8}{kW}\:. \\ $$

Commented by Tinkutara last updated on 25/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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