Chapter 4, Problem 82GP

(a)

To determine

To find: The maximum acceleration of train.

Answer to Problem 82GP

  $0.378{\text{ m/s}}^2$

Explanation of Solution

Given Data:

  $\begin{array}{l}\text{Mass of train, }m=660\text{ metric tons}\\ m=660,000\text{ kg}\\ \text{Maximum force exerted horizontally against the track,}\\ F\text{= 400 KN}\Rightarrow F{\text{ = 400 ×10}}^{3}N\\ \text{Maximum velocity, }\nu =300\text{ km/hr}\\ \nu =\frac{300\times 1000}{60\times 60}=83.33\text{ m/s}\\ \text{Exerted force, }f=150\text{ kN}\\ f=150{\text{ ×10}}^3\text{N}\end{array}$

Formula Used:

  $\text{Acceleration, }a=\frac{\text{Net force }\left({\text{F}}_{net}\right)}{\text{mass}\left(m\right)}$

Calculation:

  

  $\begin{array}{l}\text{Acceleration, }a=\frac{F_{net}}{m}\\ a=\frac{400\times {10}^3-150\times {10}^3}{660,000}\\ a=\frac{250\times {10}^3}{660,000}\\ a=0.378{\text{ m/s}}^2\end{array}$

Conclusion:

  $\text{The maximum acceleration of train is 0}{\text{.378 m/s}}^2.$

(b)

To determine

To find: The force of air resistance at tap speed.

Answer to Problem 82GP

Zero

Explanation of Solution

Given Data:

  $\begin{array}{l}\text{Mass of train, }m=660\text{ metric tons}\\ m=660000\text{ kg}\\ \text{Maximum force exerted horizontally against the trains, }F=400\text{ kN}\\ \text{Exerted force, }f=150\text{ kN}\end{array}$

Formula Used:

  $F_{net}=ma,F_{net}=\text{ Net force on the train}$

Calculation:

  $\begin{array}{l}\text{Substituting the value}\\ F_{net}=ma\\ \text{For the maximum speed net force on the train will be zero}\text{.}\\ F_{net}=0\end{array}$

Conclusion:

The force of air resistance at top speed is zero.