Chapter 18, Problem 72GP

(a)

To determine

The required horsepower.

Answer to Problem 72GP

  $4.0hp$

Explanation of Solution

Given:

Average frictional force, $F=240N$

Average speed, $v=45km/h$

Current rating in one hour, $I=95A$

Voltage supply to each battery, $V=12V$

Number of batteries, $n=24$

Mass of Car, $m=1560kg$

Formula used:

The relationship between power, force and velocity is given by:

  $P=Fv$

Where,

  $P$ is power

  $F$ is force

v is velocityCalculation:

Substitute all the given values in the formula: $P=Fv$

  $P=\left(240N\right)\left(45km/hr\right)\left(\frac{1m/s}{3.6km/hr}\right)$

  $P=3000W\left(\frac{1hp}{746W}\right)=4.0hp$

Conclusion: The required horsepower is $4.0hp$

(b)

To determine

The distance in kilometres after which the batteries must be recharged.

Answer to Problem 72GP

  $410km$

Explanation of Solution

Given:

Average frictional force, $F=240N$

Average speed, $v=45km/h$

Current rating in one hour, $I=95A$

Voltage supply to each battery, $V=12V$

Number of batteries, $n=24$

Mass of Car, $m=1560kg$

Formula used:

Formula used to calculate distance: $d=\frac{Qv}{F}$

Where,

  $d$ is distance, $Q$ is specific energy, $F$ is force, v is velocity

Calculation:

It is known that, $Q=\frac{W}{V}$ , where, $W=ItV$

Now, here $Q=\frac{ItV}{V}=It$ where, $I$ is current and $t$ is time.

  $Q=95\times 60\times 60=3.42\times {10}^{5}C$

Substitute all the given values in the formula: $d=\frac{QV}{F}$

  $d=\frac{24\left(3.42\times {10}^{5}C\right)\left(12V\right)}{240N}=410km$

Conclusion: The required distance is $410km$ .