Question

Integrated Concepts

(a) Assuming $95.0\mathrm{\%}$efficiency for the conversion of electrical power by the motor, what current must the $12.0\mathrm{V}$ batteries of a $750\mathrm{k}\mathrm{g}$ electric car be able to supply: (a) To accelerate from rest to $25.0m/\phantom{ms}s$ in $1.00\mathrm{m}\mathrm{i}\mathrm{n}$? (b) To climb a $2.00\times 10^2\mathrm{m}$ high hill in $2.00\mathrm{m}\mathrm{i}\mathrm{n}$ at a constant $25.0m/\phantom{ms}s$ speed while exerting $5.00\times 10^2\mathrm{N}$ of force to overcome air resistance and friction? (c) To travel at a constant $25.0m/\phantom{ms}s$ speed, exerting a $5.00\times 10^2\mathrm{N}$force to overcome air resistance and friction? See Figure $20.44$.

Short answer

1. The amount of current is $343\mathrm{A}$.
2. The amount of current is $2.17\mathrm{k}\mathrm{A}$.
3. The amount of current is $1.10\mathrm{k}\mathrm{A}$.

Step-by-step solution

Definition of Power:

The power of a process is the amount of some type of energy converted into a different type divided by the time interval$\mathrm{\Delta }\mathrm{t}$in which the process occurred:

$\mathrm{P}=\frac{\mathrm{\Delta }\mathrm{E}}{\mathrm{\Delta }\mathrm{t}}$ ….. (1)

The SI unit of power is the watt.

As you know, one watt is equal to one joule per second. Therefore,

$1W=1J/\phantom{Js}s$

The power of any electric circuit is defined by the formula

$\mathrm{P}=\mathrm{I}\mathrm{\Delta }\mathrm{V}$ ….. (2)

Here,$\mathrm{I}$is the current and$\mathrm{I}\mathrm{\Delta }\mathrm{V}$is the potential difference.

Definition of Kinetic Energy

The kinetic energy of any particle,

$\mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$ ….. (3)

Where$\mathrm{m}$is the mass and$\mathrm{v}$is the velocity of the particle.

Definition of Gravitational potential energy

The gravitational potential energy is given by

${\mathrm{U}}_{\mathrm{g}}=\mathrm{m}\mathrm{g}\mathrm{h}$ ….. (4)

Here,$\mathrm{m}$is the mass of the particle,$\mathrm{g}$is the acceleration due to gravity and$\mathrm{h}$is the height of the particle.

Definition of Work done:

The work done is given by

$\mathrm{W}=\mathrm{F}\mathrm{d}\mathrm{c}\mathrm{o}\mathrm{s}\theta$ ….. (5)

Here,$\mathrm{F}$is the exerted force,$\mathrm{d}$is the displacement,$\theta$is the angle between$\overrightarrow{\mathrm{F}}$and$\overrightarrow{\mathrm{d}}$.

Definition of average speed:

The average speed is

$\mathrm{v}=\frac{\mathrm{\Delta }\mathrm{x}}{\mathrm{t}}......(6)$

Where$\mathrm{\Delta }\mathrm{x}$is the change in displacement or position and$\mathrm{t}$is the time.

The given data

- The efficiency for conversion of electrical power by the motor is:$\eta =95.0\mathrm{\%}$.

- The potential difference across the batteries:$\mathrm{\Delta }\mathrm{V}=12.0\mathrm{V}$.

- The mass of the electric car is:$\mathrm{m}=750\mathrm{k}\mathrm{g}$.

(a) To accelerate the calculation of current

The initial speed of the car is:${\mathrm{v}}_{\mathrm{i}}=0$.

The final speed of the car is: ${\mathrm{v}}_{\mathrm{f}}=25.0\mathrm{m}/\mathrm{s}$.

The time interval of the acceleration is:

$\begin{array}{rl}\mathrm{\Delta }\mathrm{t}& =\left(1.00\mathrm{m}\mathrm{i}\mathrm{n}\right)\left(\frac{60\mathrm{s}}{1\mathrm{m}\mathrm{i}\mathrm{n}}\right)\\ & =60.0\mathrm{s}\end{array}$

Given that the efficiency of the conversion is$95.0\mathrm{\%}$, so:

$\left(0.95\right){\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\mathrm{\Delta }\mathrm{K}$

Substitute $\mathrm{P}\mathrm{\Delta }\mathrm{t}$ for ${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$ from Equation (1) and the change in kinetic energy of the car is from Equation (3):

$\left(0.95\right)\mathrm{P}\mathrm{\Delta }\mathrm{t}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{f}}}^2-\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{i}}}^2$

Substitute $\mathrm{I}\mathrm{\Delta }\mathrm{V}$ for $\mathrm{P}$ in the above equation.

$\left(0.95\right)\left(\mathrm{I}\mathrm{\Delta }\mathrm{V}\right)\mathrm{\Delta }\mathrm{t}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{f}}}^2-\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{i}}}^2$

Rearrange the above equation for current as below.

$\mathrm{I}=\frac{{\mathrm{m}\mathrm{v}}_{\mathrm{f}}^2}{2\left(0.95\right)\mathrm{\Delta }\mathrm{V}\mathrm{\Delta }\mathrm{t}}$

Solve the above equation by substituting known numerical values in the above equation.

$\begin{array}{rl}\mathrm{I}& =\frac{\left(750\mathrm{k}\mathrm{g}\right){\left(25.0m/\phantom{ms}s\right)}^2}{2\left(0.95\right)\left(12.0\mathrm{V}\right)\left(60.0\mathrm{s}\right)}\\ & =343\mathrm{A}\end{array}$

Therefore, in this case the amount of current is $343\mathrm{A}$.

(b) To climb the calculation of current

The change in height of the car is: $\mathrm{\Delta }\mathrm{y}=2.00\times 10^2\mathrm{m}$.

The time interval of climbing is:

$\begin{array}{rl}\mathrm{\Delta }\mathrm{t}& =\left(2.00\mathrm{m}\mathrm{i}\mathrm{n}\right)\left(\frac{60\mathrm{s}}{1\mathrm{m}\mathrm{i}\mathrm{n}}\right)\\ & =120\mathrm{s}\end{array}$

The speed of the car, $\mathrm{v}=25.0m/\phantom{ms}s$.

The force exerted by the car to overcome air resistance and friction,$\mathrm{F}=5.00\times 10^2\mathrm{N}$

The electrical energy ${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$supplied by the battery in this case goes into increasing only the gravitational potential energy ${\mathrm{U}}_{\mathrm{g}}$ of the car and exerting a positive work $\mathrm{W}$ to overcome air resistance and friction.

The kinetic energy of the car is constant because the car is moving with constant speed. Given efficiency of the conversion is $95.0\mathrm{\%}$.So:

$\left(0.95\right){\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}=\mathrm{\Delta }{\mathrm{U}}_{\mathrm{g}}+\mathrm{W}$

Substitute $\mathrm{I}\mathrm{\Delta }\mathrm{V}\mathrm{\Delta }\mathrm{t}$ for ${\mathrm{E}}_{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}$ and $\mathrm{m}\mathrm{g}\mathrm{\Delta }\mathrm{y}$ for $\mathrm{\Delta }{\mathrm{U}}_{\mathrm{g}}$, and $\mathrm{F}\mathrm{d}$ for $\mathrm{W}$ in the above equation.

$\left(0.95\right)\mathrm{I}\mathrm{\Delta }\mathrm{V}\mathrm{\Delta }\mathrm{t}=\mathrm{m}\mathrm{g}\mathrm{\Delta }\mathrm{y}+\mathrm{F}\mathrm{d}$

Here, $\mathrm{F}$ is the force exerted by the car and $\mathrm{d}$ is the distance moved by the car.

$\left(0.95\right)I\mathrm{\Delta }V\mathrm{\Delta }t=mg\mathrm{\Delta }y+Fv\mathrm{\Delta }t$

Rearrange the above equation for current $\mathrm{I}$ as below.

$I=\frac{mg\mathrm{\Delta }y+Fv\mathrm{\Delta }t}{(0.95)\mathrm{\Delta }V\mathrm{\Delta }t}$

Substitute numerical values:

$\begin{array}{rl}I& =\frac{(750\mathrm{k}\mathrm{g}\left(9.80m/\phantom{m{s}^2}{s}^2\right)\left(2.00\times {10}^{2}m\right)+\left(5.00\times {10}^{2}N\right)(25.0m/\phantom{ms}s)(120s)}{0.95\times 12.0V\times 120s}\\ & =2.17\times {10}^{3}A\\ & =\left(2.17\times {10}^{3}A\right)\left(\frac{1kA}{1000A}\right)\\ & =2.17kA\end{array}$

Therefore, in this case the amount of current is $2.17kA$.

(c) To travel the calculation of current:

The speed of the car is $\mathrm{v}=25.0m/\phantom{ms}s$.

The force exerted by the car to overcome air resistance and friction is,$\mathrm{F}=5.00\times 10^2\mathrm{N}$.

Overcome air resistance and friction. Given that efficiency of the conversion is$95.0\mathrm{\%}$.

Therefore,

$\left(0.95\right)E_{electrical}=W$

Here, $\mathrm{W}$ is found from Equation (5):

$\left(0.95\right)I\mathrm{\Delta }V\mathrm{\Delta }t=Fd$

Here, $\mathrm{F}$ is the force exerted by the car and $\mathrm{d}$ is the distance moved by the car found from Equation (6):

$\begin{array}{r}l\left(0.95\right)I\mathrm{\Delta }V\mathrm{\Delta }t=Fv\mathrm{\Delta }t\\ \left(0.95\right)I\mathrm{\Delta }V=Fv\end{array}$

Solve the above equation for current $\mathrm{I}$ as below.

$I=\frac{Fv}{\left(0.95\right)\mathrm{\Delta }V}$

Substitute known numerical values in the above expression.

$\begin{array}{rl}I& =\frac{\left(5.00\times {10}^{2}N\right)(25.0m/\phantom{ms}s)}{0.95\times 12.0V}\\ & =1.10\times {10}^{3}A\\ & =\left(1.10\times {10}^{3}A\right)\left(\frac{1kA}{1000A}\right)\\ & =1.10kA\end{array}$

Therefore, in this case the amount of current is $1.10kA$.