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

$$\mathrm{A}\mathrm{train}\mathrm{weighing}100\mathrm{metric}\mathrm{ton}\mathrm{is}$$$$\mathrm{running}\mathrm{on}\mathrm{a}\mathrm{level}\mathrm{track}\mathrm{with}\mathrm{a}\mathrm{uniform}$$$$\mathrm{speed}\mathrm{of}72\mathrm{km}{\mathrm{h}}^{-1}.\mathrm{If}\mathrm{the}\mathrm{frictional}$$$$\mathrm{resistance}\mathrm{amounts}\mathrm{to}0.5\mathrm{kg}\mathrm{per}\mathrm{metric}$$$$\mathrm{ton},\mathrm{find}\mathrm{the}\mathrm{power}\mathrm{of}\mathrm{the}\mathrm{engine}.$$

Answered by ajfour last updated on 24/Oct/17

$$letmetricton=M$$$$resistanceof0.5kg/Mmeans$$$$resistanceequalto(weightof$$$$0.5kg)/M=(0.5kg\times 9.8m/s^2)/M$$$$=4.9N/M$$$$netresistance=\frac{4.9N}{M}\times 100M$$$$=490N$$$$P=Fv=490N\times \left(\frac{5}{18}\times \frac{72m}{s}\right)$$$$=9800W=9.8kW.$$

Commented by Tinkutara last updated on 25/Oct/17

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

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