Question

A battery-operated car uses a 12.0-V system. Find the charge the batteries must be able to move in order to accelerate the 956kg car from rest to 158m/s.

Answer 1

Answer:

1.99 x 10⁶C

Explanation:

The electrical energy ($E_{E}$) stored in the battery will give the car some kinetic energy ($E_{K}$) which will cause the car to move from rest to some other point.

i.e

$E_{E}$ = $E_{K}$                 ------------------(i)

But;

$E_{E}$ =  $\frac{1}{2}$ x Q x V;            -------------------(ii)

Where;

Q = charge on the battery

V = potential difference or voltage of the battery = 12.0V

Also

$E_{K}$ = ($\frac{1}{2}$ x m x v²) - ($\frac{1}{2}$ x m x u²)            -----------------(iii)

Where;

m = mass of the car = 956kg

v = final velocity of the car = 158m/s

u = initial velocity of the car = 0   [since the car starts from rest]

Substitute equations (ii) and (iii) into equation (i) as follows;

$\frac{1}{2}$ x Q x V = ($\frac{1}{2}$ x m x v²) - ($\frac{1}{2}$ x m x u²)       -----------------(iv)

Substitute all necessary values into equation (iv) as follows;

$\frac{1}{2}$ x Q x 12.0 = ($\frac{1}{2}$ x 956 x 158²) - ($\frac{1}{2}$ x 956 x 0²)

$\frac{1}{2}$ x Q x 12.0 = ($\frac{1}{2}$ x 956 x 158²) - (0)

$\frac{1}{2}$ x Q x 12.0 = ($\frac{1}{2}$ x 956 x 158²)

$\frac{1}{2}$ x Q x 12.0 = 11932792

6Q = 11932792

Solve for Q;

Q =  11932792 / 6

Q = 1988798.67 C

Q = 1.99 x 10⁶C

Therefore, the amount of charge the batteries must have is 1.99 x 10⁶C