Question

Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 708-kg electric car be able to supply to do the following?

(a) accelerate from rest to 25.0 m/s in 1.00 min
1 (answer in Amps)

(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 423 N of force to overcome air resistance and friction

Answers

  1. Answer:

    (a). The current is 323.4 A.

    (b). The current is 1942 A.

    Explanation:

    Given that,

    Efficiency = 95.0 %

    Voltage = 12.0 V

    Mass of electric car= 708 Kg

    Height = 200 m

    We need to calculate the change in kinetic energy

    Using formula of kinetic energy

    \Delta K=K_{f}-K_{i}

    Put the value into the formula

    \Delta K=\dfrac{1}{2}mv^2-0

    \Delta K=\dfrac{1}{2}\times708\times(25)^2-0

    \Delta K=221250\ J

    We need to calculate the output power

    Using formula of power

    P_{o}=\dfrac{\Delta K}{t}

    P_{o}=\dfrac{221250}{60}

    P_{o}=3687.5\ W

    We need to calculate the current

    Using formula of electric power

    P_{in}=iV

    P_{o}=0.95P_{in}

    P_{0}=0.95\times iV

    Put the value into the formula

    3687.5=0.95\times i\times12.0

    i=\dfrac{3687.5}{0.95\times12.0}

    i=323.4\ A

    The current is 323.4 A.

    (b). We need to calculate the distance

    d=vt

    Put the value into the formula

    d=25\times2\times60

    d=3000\ m

    We need to calculate the force

    Using formula of force

    F=mg\sin\theta

    Put the value into the formula

    F=708\times9.8\times\dfrac{200}{3000}

    F=462.56\ N

    We need to calculate the power

    Using formula of power

    P=F\times v

    Put the value into the formula

    P=(462.56+423)\times25

    P=22139\ W

    We need to calculate the current

    Using formula of current

    I=\dfrac{P}{V}

    Put the value into the formula

    I=\dfrac{22139\times100}{95\times12}

    I=1942.0\ A

    The current is 1942 A.

    Hence, (a). The current is 323.4 A.

    (b). The current is 1942 A.

  2. Answer:

    a) I=646.9298\ A

    b) I=1942.018\ A

    Explanation:

    Given:

    • efficiency of the motor, \eta=0.95
    • voltage of the battery, V=12\ V
    • mass of the car, m=708\ kg

    a)

    initial velocity, u=0\ m.s^{-1}

    final velocity, v=25\ m.s^{-1}

    time taken for the acceleration, t=1\ min=60\ s

    Now we know by the Newton’s second law of motion:

    F=m.a

    F=708\times \frac{(25-0)}{60}

    F=295\ N

    Now the power will be :

    P=F.v

    P=295\times 25

    P=7375\ W

    According to the question:

    0.95 times of the electrical power should yield this mechanical power.

    P=V.I\times 0.95

    7375=12\times I\times 0.95

    I=\frac{7375}{12\times 0.95}

    I=646.9298\ A

    b)

    height climbed by the car, h=200\ m

    velocity of climb, v=25\ m.s^{-1}

    time taken to climb the height, t=2\ min=120\ s

    force exerted to overcome air and frictional resistances, f=423\ N

    Now the Power required to climb the hill:

    P=\frac{m.g.h}{t} +f\times v

    P=\frac{708\times 9.8\times 200}{120}+423\times 25

    P=22139\ W

    Now according to the electrical efficiency:

    P=V.I\times 0.95

    22139=12\times I\times 0.95

    I=1942.018\ A

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