A 1400 kg car is traveling at 24 m/s when the driver takes his foot off of the gas. The car eventually rolls to a stop after 225 m. What the force of friction acting on the car?

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Answer 1 · 2017-02-25T06:06:18

$1792N$

In this case, the only force is friction.

So $F = 1400 * a$

Lets pause for a moment to find $a$ .

Let's use one of the projectile motion equations. We choose this one because it has all the variable we know except for $a$ so we can solve for $a$ .

$V_f^2 = V_i^2 + 2ad$
$0 = 576 + 2*a*225$
$a = -1.28$ (indicating the negative direction)

So back to our original equation:

$F = 1400 * a$
$F = 1400 * -1.28$
$F = 1400 * -1.28$
$=1792N$ Opposite the direction of the movement of the car