Question

A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force for an hour.

Answer 1

Answer:

3894531 coulombs

Explanation:

1 hour = 3600 seconds

Let g = 10m/s2

The distance that the car travel at 25 m/s over an hour is

s = 25 * 3600 = 90000 m

The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:

$\sum E = E_k + E_p + E_W$

$\sum E = mv^2/2 + mgh + Fs$

$\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J$

This energy is drawn from the battery over an hour (3600 seconds), so its power must be

$P = E / t = 46734375/3600 = 12982 W$

The system is 12V so its current is

$I = P/U = 12982 / 12 = 1081.8 A$ or 1081.8 Coulombs/s

The the total charge it needs for 1 hour (3600 s) is

C = 1081.8 * 3600 = 3894531 coulombs

Answer 2

The quantity of charge the batteries must be able to move is equal to 3.9 μC.

Given the following data:

• Voltage = 12.0 Volts

• Mass = 750 kg

• Speed = 25.0 m/s

• Height = 200 meters.

• Force = 500 Newton.

• Time = 1 hour = 3600 seconds.

Scientific data:

• Acceleration due to gravity = 9.8 $m/s^2$

To find the quantity of charge the batteries must be able to move:

In this scenario, we would calculate the distance traveled and the total energy that is possessed by this battery-operated car.

For the distance.

Mathematically, the distance covered by an object is given by this formula:

$Distance = speed \times time\\ \\ Distance = 25 \times 3600$

Distance = 90,000 meters.

For the total energy:

$E = mgh + \frac{1}{2} mv^2 + Fd\\ \\ E=[750\times 9.8 \times 200] + \frac{1}{2} \times 750 \times 25^2 + [500 \times 90000]\\ \\ E=1470000+234375+45000000$

Total energy = 46,704,375 Joules.

Next, we would calculate the power consumed by this battery-operated car:

$Power = \frac{Energy}{time}\\ \\ Power = \frac{46,704,375}{3600}$

Power = 12,973.44 Watts.

Also, we would calculate the current:

$Current = \frac{power}{voltage} \\ \\ Current = \frac{12,973.44}{12}$

Current = 1,081.12 Amperes.

Now, we can calculate the quantity of charge the batteries must be able to move:

$Q = current \times time\\ \\ Q = 1081.12 \times 3600$

Q = 3,892,032 Coulombs.

Note: 1 μC = $1 \times 10^6 \;C$

Q = 3.9 μC

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