Question
It takes a force of 53 $\mathrm{kN}$ on the lead car of a 16 -car passenger train with mass $9.1 \times 10^{5} \mathrm{kg}$ to pull it at a constant 45 $\mathrm{m} / \mathrm{s}$ $(101 \mathrm{mi} / \mathrm{h})$ on level tracks. (a) What power must the locomotive $(101 \mathrm{mide} \text { to the lead car? (b) How much more power to the lead car }$ than calculated in part (a) would be needed to give the train an up a 1.5$\%$ grade (slope angle $\alpha=\arctan 0.015$ ) at a constant 45 $\mathrm{m} / \mathrm{s} ?$
Step 1
In this case, the force is 53,000 N and the velocity is 45 m/s. Therefore, the power is given by: \[ P = F \cdot v = 53,000 \, \text{N} \cdot 45 \, \text{m/s} = 2.4 \times 10^{6} \, \text{Watt} \] Show more…
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It takes a force of 53 $\mathrm{kN}$ on the lead car of a 16 -car passenger train with mass $9.1 \times 10^{5} \mathrm{kg}$ to pull it at a constant 45 $\mathrm{m} / \mathrm{s}$ $(101 \mathrm{mi} / \mathrm{h})$ on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in part (a) would be needed to give the train an acceleration of 1.5 $\mathrm{m} / \mathrm{s}^{2}$ , at the instant that the train has a speed of 45 $\mathrm{m} / \mathrm{s}$ on level tracks? (c) How much more power to the lead car than that calculated in part (a) would be needed to move the train up a 1.5$\%$ grade (slope angle $\alpha=\arctan 0.015 )$ at a constant 45 $\mathrm{m} / \mathrm{s} ?$
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