The breaking acceleration for a car is $-12.2\frac{m}{s^2}$. The car has a mass of 925kg. I want to find the frictional force. Here is what I have so far:

$$\Sigma\vec{F} = m\vec{a}$$ $$\Sigma\vec{F} = (925kg)(-12.2m/s^2)$$ $$\Sigma\vec{F} = -11285N$$

So now I have the net force of the car. Would this be the same as the frictional force (any action has an equal and opposite reaction)? Or do I need to use the Friction Law formula? ${}{}$



1 Answer 1


From the question you posted, one can not tell whether the frictional force is the sole force acting on the car. But, it is safe to assume that when someone is braking, that is the only significant force acting on it.

So, the net force is equal to the frictional force.

  • 1
    $\begingroup$ Just to be irritatingly correct, the frictional force is not the only force on the car; a gravitational and normal force will be present. Only by assuming that the surface is horizontal can you say the net force is equal to the frictional force. $\endgroup$
    – BMS
    Commented Jan 27, 2014 at 0:52
  • $\begingroup$ I completely agree with you, the way the OQ is worded required me to make significant assumptions, including that the surface is horizontal. In saying that it is the only significant force, I meant that it is the only force contributing to the measured acceleration. Definitely could have worded it a bit better. $\endgroup$
    – Richard P
    Commented Jan 27, 2014 at 0:54

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