We have to do a little work with Netwon's 2nd. For both cars I will use a coordinate system with +y upward.Originally Posted by LLELLA
Big car (BC): The free body diagram has a weight wBC downward, a normal force NBC upward, and a friction force, say, to the left. (I am taking the car to be moving to the right. The choice doesn't matter.) So in the y direction:
Thus NBC = (mBC)g. Thus the friction force on BC is fBC = (mu)NBC = (mu)(mBC)g = 2(mu)mg. (mBC = 2m)
Now the work done by friction to stop the car will be equal in magnitude to the change in kinetic energy of the car. Specifically:
(The work done by friction is always negative, do the dot product if you aren't convinced. Here vBC is the initial speed of the BC, dBC is the stopping distance, and the final speed is, of course, zero.)
In terms of the given variables: mBC = 2m, vBC = v
Small car: Virtually identical to the BC, except now mSC = m. So fSC = (mu)mg. So using the work formula once again, and putting mSC = m and vSC = 2v:
So the small car takes 4x the distance to stop.