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