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

Answer 1

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:

$P=V.I\times 0.95$

$22139=12\times I\times 0.95$

$I=1942.018\ A$

Answer 2

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.