Question

Integrated Concepts

(a) Assuming \({\rm{95}}{\rm{.0\% }}\)efficiency for the conversion of electrical power by the motor, what current must the \({\rm{12}}{\rm{.0 V}}\) batteries of a \({\rm{750 kg}}\) electric car be able to supply: (a) To accelerate from rest to \({\rm{25}}{\rm{.0\; }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}\) in \({\rm{1}}{\rm{.00\;min}}\)? (b) To climb a \({\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{ m}}\) high hill in \({\rm{2}}{\rm{.00 min}}\) at a constant \({\rm{25}}{\rm{.0 }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}\) speed while exerting \({\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{ N}}\) of force to overcome air resistance and friction? (c) To travel at a constant \({\rm{25}}{\rm{.0 }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}\) speed, exerting a \({\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;N}}\)force to overcome air resistance and friction? See Figure \({\rm{20}}{\rm{.44}}\).

Short answer

1. The amount of current is \({\rm{343\;A}}\).
2. The amount of current is \({\rm{2}}{\rm{.17\;kA}}\).
3. The amount of current is \({\rm{1}}{\rm{.10kA}}\).

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\({\rm{\Delta t}}\)in which the process occurred:

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

The SI unit of power is the watt.

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

\(1{\rm{ }}W = 1{\rm{ }}{J \mathord{\left/ {\vphantom {J s}} \right. \ } s}\)

The power of any electric circuit is defined by the formula

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

Here,\({\rm{I}}\)is the current and\({\rm{I\Delta V}}\)is the potential difference.

Definition of Kinetic Energy

The kinetic energy of any particle,

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

Where\({\rm{m}}\)is the mass and\({\rm{v}}\)is the velocity of the particle.

Definition of Gravitational potential energy

The gravitational potential energy is given by

\({{\rm{U}}_{\rm{g}}} = {\rm{mgh}}\) ….. (4)

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

Definition of Work done:

The work done is given by

\({\rm{W}} = {\rm{Fdcos\theta }}\) ….. (5)

Here,\({\rm{F}}\)is the exerted force,\({\rm{d}}\)is the displacement,\({\rm{\theta }}\)is the angle between\({\rm{\vec F}}\)and\({\rm{\vec d}}\).

Definition of average speed:

The average speed is

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

Where\({\rm{\Delta x}}\)is the change in displacement or position and\({\rm{t}}\)is the time.

The given data

- The efficiency for conversion of electrical power by the motor is:\({\rm{\eta = 95}}{\rm{.0\% }}\).

- The potential difference across the batteries:\({\rm{\Delta V = 12}}{\rm{.0\;V}}\).

- The mass of the electric car is:\({\rm{m = 750\;kg}}\).

(a) To accelerate the calculation of current

The initial speed of the car is:\({{\rm{v}}_{\rm{i}}}{\rm{ = 0}}\).

The final speed of the car is: \({{\rm{v}}_{\rm{f}}}{\rm{ = 25}}{\rm{.0\;m/s}}\).

The time interval of the acceleration is:

\(\begin{align}{\rm{\Delta t}} &= \left( {{\rm{1}}{\rm{.00\;min}}} \right)\left( {\frac{{{\rm{60\;s}}}}{{{\rm{1\;min}}}}} \right)\\ &= {\rm{60}}{\rm{.0\;s}}\end{align}\)

Given that the efficiency of the conversion is\({\rm{95}}{\rm{.0\% }}\), so:

\(\left( {{\rm{0}}{\rm{.95}}} \right){{\rm{E}}_{{\rm{electrical }}}} = {\rm{\Delta K}}\)

Substitute \({\rm{P\Delta t}}\) for \({{\rm{E}}_{{\rm{electrical }}}}\) from Equation (1) and the change in kinetic energy of the car is from Equation (3):

\(\left( {{\rm{0}}{\rm{.95}}} \right){\rm{P\Delta t}} = \frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}_{\rm{f}}}^2 - \frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{v}}_{\rm{i}}}^2\)

Substitute \({\rm{I\Delta V}}\) for \({\rm{P}}\) in the above equation.

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

Rearrange the above equation for current as below.

\({\rm{I}} = \frac{{{\rm{mv}}_{\rm{f}}^{\rm{2}}}}{{{\rm{2}}\left( {{\rm{0}}{\rm{.95}}} \right){\rm{\Delta V\Delta t}}}}\)

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

\(\begin{align}{\rm{I}} &= \frac{{\left( {{\rm{750\;kg}}} \right){{\left( {{\rm{25}}{\rm{.0 }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}} \right)}^{\rm{2}}}}}{{{\rm{2}}\left( {{\rm{0}}{\rm{.95}}} \right)\left( {{\rm{12}}{\rm{.0\;V}}} \right)\left( {{\rm{60}}{\rm{.0\;s}}} \right)}}\\ &= {\rm{343\;A}}\end{align}\)

Therefore, in this case the amount of current is \({\rm{343\;A}}\).

(b) To climb the calculation of current

The change in height of the car is: \({\rm{\Delta y}} = {\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;m}}\).

The time interval of climbing is:

\(\begin{align}{\rm{\Delta t}} &= \left( {{\rm{2}}{\rm{.00\;min}}} \right)\left( {\frac{{{\rm{60\;s}}}}{{{\rm{1\;min}}}}} \right)\\ &= {\rm{120\;s}}\end{align}\)

The speed of the car, \({\rm{v}} = {\rm{25}}{\rm{.0 }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}\).

The force exerted by the car to overcome air resistance and friction,\({\rm{F}} = {\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;N}}\)

The electrical energy \({{\rm{E}}_{{\rm{electrical}}}}\)supplied by the battery in this case goes into increasing only the gravitational potential energy \({{\rm{U}}_{\rm{g}}}\) of the car and exerting a positive work \({\rm{W}}\) 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 \({\rm{95}}{\rm{.0\% }}\).So:

\(\left( {{\rm{0}}{\rm{.95}}} \right){{\rm{E}}_{{\rm{electrical }}}} = {\rm{\Delta }}{{\rm{U}}_{\rm{g}}}{\rm{ + W}}\)

Substitute \({\rm{I\Delta V\Delta t}}\) for \({{\rm{E}}_{{\rm{electical}}}}\) and \({\rm{mg\Delta y}}\) for \({\rm{\Delta }}{{\rm{U}}_{\rm{g}}}\), and \({\rm{Fd}}\) for \({\rm{W}}\) in the above equation.

\(\left( {{\rm{0}}{\rm{.95}}} \right){\rm{I\Delta V\Delta t}} = {\rm{mg\Delta y}} + {\rm{Fd}}\)

Here, \({\rm{F}}\) is the force exerted by the car and \({\rm{d}}\) is the distance moved by the car.

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

Rearrange the above equation for current \({\rm{I}}\) as below.

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

Substitute numerical values:

\(\begin{align}I &= \frac{{\left( {750\;{\rm{ kg}}\left( {9.80{\rm{ }}{m \mathord{\left/ {\vphantom {m {{s^2}}}} \right. \ } {{s^2}}}} \right)\left( {2.00 \times {{10}^2}\;{\rm{ }}m} \right) + \left( {5.00 \times {{10}^2}{\rm{ }}\;N} \right)(25.0\;{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s})(120{\rm{ }}s)} \right.}}{{0.95 \times 12.0{\rm{ }}V \times 120{\rm{ }}s}}\\ &= 2.17 \times {10^3}{\rm{ }}A\\ &= \left( {2.17 \times {{10}^3}{\rm{ }}A} \right)\left( {\frac{{1{\rm{ }}kA}}{{1000{\rm{ }}A}}} \right)\\ &= 2.17{\rm{ }}kA\end{align}\)

Therefore, in this case the amount of current is \(2.17{\rm{ }}kA\).

(c) To travel the calculation of current:

The speed of the car is \({\rm{v}} = {\rm{25}}{\rm{.0 }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s}\).

The force exerted by the car to overcome air resistance and friction is,\({\rm{F}} = {\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;N}}\).

Overcome air resistance and friction. Given that efficiency of the conversion is\({\rm{95}}{\rm{.0\% }}\).

Therefore,

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

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

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

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

\(\begin{align}{l}\left( {0.95} \right)I\Delta V\Delta t = Fv\Delta t\\\left( {0.95} \right)I\Delta V = Fv\end{align}\)

Solve the above equation for current \({\rm{I}}\) as below.

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

Substitute known numerical values in the above expression.

\(\begin{align}I &= \frac{{\left( {5.00 \times {{10}^2}\;{\rm{ }}N} \right)(25.0{\rm{ }}{m \mathord{\left/ {\vphantom {m s}} \right. \ } s})}}{{0.95 \times 12.0{\rm{ }}V}}\\ &= 1.10 \times {10^3}\;A\\ &= \left( {1.10 \times {{10}^3}\;A} \right)\left( {\frac{{1{\rm{ }}kA}}{{1000\;A}}} \right)\\ &= 1.10{\rm{ }}kA\end{align}\)

Therefore, in this case the amount of current is \(1.10{\rm{ }}kA\).