Chapter 25, Problem 67P

(a)

To determine

Minimum power delivered by the electric motor

Answer to Problem 67P

  $26640\text{ W}$

Explanation of Solution

Given:

Average force due to air drag and rolling friction

  $\begin{array}{l}=F=1.20\text{ kN }\\ \text{= 1}\text{.20 (1000) N }\\ \text{= 1200 N}\end{array}$

Speed of the car

  $\begin{array}{l}=v=80\frac{\text{km}}{\text{h}}\\ =80\frac{\text{km}}{\text{h}}\left(\frac{\text{1000 m}}{\text{1km}}\right)\left(\frac{\text{1h}}{\text{3600 s}}\right)\\ =22.2\frac{\text{m}}{\text{s}}\end{array}$

Minimum power delivered by the battery $=P_{battery}$

Formula Used:

Power in terms of Voltage “F” and constant velocity“v” is given as

  $P=Fv$

Calculation:

Minimum power delivered by the battery is:

  $\begin{array}{l}{P}_{battery}=Fv\\ P_{battery}=(1200)(22.2)\\ P_{battery}=26640\text{ W}\end{array}$

Conclusion:

Hence, minimum power delivered by the battery is $26640\text{ W}$ .

(b)

To determine

Total charge delivered by series combination of ten batteries.

Answer to Problem 67P

  $\text{5}\text{.76}\times {\text{10}}^5\text{ C}$

Explanation of Solution

Given:

Charge delivered by each battery $=Q=160\text{ A}\text{.h = }160(3600)\text{ A}\text{.s = 5}\text{.76}\times {\text{10}}^5\text{ C}$

Total charge delivered by series combination of batteries $=Q_s$

Formula Used:

Charge in terms of current “I” and “t” is given as

  $Q=It$

Calculation:

Power delivered by single battery is:

  $\begin{array}{l}Q=160\text{ A}\cdot \text{h }\\ \text{Q= }160(3600)\text{ A}\cdot \text{s }\\ \text{Q = 5}\text{.76}\times {\text{10}}^5\text{ C}\end{array}$

In series charge remains same. Hence the charge for series combination of ten batteries is same as the charge delivered by a single battery.

Hence

  ${\text{Q}}_s\text{= Q = 5}\text{.76}\times {\text{10}}^5\text{ C}$

Conclusion:

Hence, thecharge delivered by series combinationis $\text{5}\text{.76}\times {\text{10}}^5\text{ C}$ .

(c)

To determine

Total energy delivered by series combination of ten batteries.

Answer to Problem 67P

  $\text{6}\text{.91}\times {\text{10}}^7\text{ J}$

Explanation of Solution

Given:

Charge delivered by each battery $=Q=160\text{ A}\text{.h = }160(3600)\text{ A}\text{.s = 5}\text{.76}\times {\text{10}}^5\text{ C}$

EMF of each battery $=E=12\text{ volts}$

Number of batteries in series $=n=10$

Total EMF provided by series combination of batteries $=E_s$

Total energy delivered by the combination of batteries $=U_s$

Formula Used:

Energy provided by a battery of EMF “V” by delivering a charge “Q” is given as

  $U=QE$

Calculation:

Total EMF provided by series combination of batteries:

  $\begin{array}{l}{E}_s=nE\\ E_s=(10)(12)\\ E_s=120\text{ volts}\end{array}$

Total energy delivered by the combination of batteries:

  $\begin{array}{l}{U}_s=Q{E}_s\\ U_s=(\text{5}\text{.76}\times {\text{10}}^5)(120)\\ U_s=\text{6}\text{.91}\times {\text{10}}^7\text{ J}\end{array}$

Conclusion:

Hence, thetotal energy delivered by the combination of batteriesis $\text{6}\text{.91}\times {\text{10}}^7\text{ J}$ .

(d)

To determine

The distance traveled by the car before recharge is required.

Answer to Problem 67P

  $5.76\times {\text{10}}^4\text{ m}$

Explanation of Solution

Given:

Distance traveled by car $=d$

Work done by force on the car $=W$

Total energy delivered by the combination of batteries $=U_s=\text{6}\text{.91}\times {\text{10}}^7\text{ J}$

Formula Used:

Work done is given as

  $W=Fd$

Where, F is the applied force and d is the distance.

Calculation:

Work done by force on the car is given as

  $W=Fd$

Using conservation of energy:

  $\begin{array}{l}W=U_s\\ Fd=\text{6}\text{.91}\times {\text{10}}^7\\ d(1200)=\text{6}\text{.91}\times {\text{10}}^7\\ d=5.76\times {\text{10}}^4\text{ m}\end{array}$

Conclusion:

Hence, thedistance traveled by caris $5.76\times {\text{10}}^4\text{ m}$ .

(e)

To determine

The cost per kilometer

Answer to Problem 67P

  $\text{\$}0.03{\text{ km}}^{-1}$

Explanation of Solution

Given:

Charge delivered by each battery $=Q=160\text{ A}\text{.h }$

Total EMF provided by series combination of batteries $=E_s=120\text{ volts}$

Cost per kilowatt-hour $=c=9\frac{\text{cents}}{\text{kW}\cdot \text{h}}$

Total cost $=C$

Total distance traveled $=d=5.76\times {10}^4\text{ m}$

Total energy delivered by the combination of batteries $=U_s$

Formula Used:

Energy provided by a battery of EMF “V” by delivering a charge “Q” is given as

  $U=QE$

Total cost is given as

  $\text{Total cost = (Cost per kilowatt-hour)(total energy)}$

Cost per unit distance is given as

  $\text{Cost per kilo-meter =}\frac{\text{Total Cost}}{\text{distance}}$

Calculation:

Total energy delivered by the combination of batteries is given as

  $\begin{array}{l}{U}_s=Q{E}_s\\ U_s=(160)(120)\\ U_s=19200\text{ kW}\cdot \text{h}\end{array}$

Total cost is given as

  $\begin{array}{l}\text{C = c}{U}_s\\ C=(9)(19200)\\ C=172800\text{ cents}\\ C=\$1728.00\end{array}$

Cost per unit distance is given as

  $\begin{array}{l}\text{Cost per kilo-meter =}\frac{\text{Total Cost}}{\text{distance}}\\ \text{Cost per kilo-meter =}\frac{\text{\$1728}}{5.76\times {10}^4}\\ \text{Cost per kilo-meter =\$}0.03{\text{ km}}^{-1}\end{array}$

Conclusion:

Hence total cost per kilometer comes out to be $\text{\$}0.03{\text{ km}}^{-1}$