question_answer

A railway engine weighing 40 metric ton is travelling along a level track at a speed of 54 km/h. What additional power is required to maintain the same speed up an incline rising 1 in 49, given \[\mu =0.1\], \[g=9.8m{{s}^{-2}}\].

Answer:

                Here,    \[m=40\,\]metric ton = \[40\times {{10}^{3}}kg\]                                 \[\upsilon =54km\,{{h}^{-1}}=\frac{54\times 1000}{60\times 60}=15\,m\,{{s}^{-1}}\] \[\sin \theta =\frac{1}{49}\]; \[\mu =0.1\]; \[g=10m{{s}^{-2}}\] Now, \[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{\left( 1/49 \right)}^{2}}}\approx 1\] Power required on level track, \[{{P}_{1}}=F\times \upsilon =\mu \,mg\times \upsilon \] Power required up an incline, \[{{P}_{2}}=\left( mg\,\sin \theta +\mu \,mg\,\cos \theta  \right)\upsilon \] Additional power required \[P={{P}_{2}}-{{P}_{1}}=\left[ mg\,\sin \theta +\mu \,mg\,\cos \theta -\mu mg \right]\upsilon \] \[=\left[ mg\,\sin \theta +\mu mg\times 1-\mu mg \right]\upsilon =mg\,\sin \theta \times \upsilon \] \[P=40\times {{10}^{3}}\times 9.8\times \frac{1}{49}\times 15=120\times {{10}^{3}}watt=120kW\]