A 1000 kg car is traveling at 28 m/s. The driver wishes to stop in a distance of 105 m. What friction between the tires and road is necessary to do...

Hello,
let's solve this problem!

So we have a `1000 kg`car travelling at `28 m/s`and
we wish to make it stop after braking for a total distance of `105 m` . We want to know what the
magnitude of the force of friction must be such that this scenario is possible.

Keep in mind we're assuming that the friction between the tires and the
road is constant along the distance.

Our best tool to solve
this problem is the concept of mechanical energy. Our car has a total kinetic energy
of:

`E = (1)/(2) mv^2`

where `m` is the mass of the car,
`1000 kg` , and `v` is its speed (in meters per second), that is, `28 m/s` . Plugging in the
values we get a total kinetic energy of

`E = (1)/(2) (1000 kg) (28 m/s)^2 =
392000 J`

The second concept that we use is that of the work `W` exerted by
a constant force `F` . To make our car stop, we have to apply a certain constant force `F` along
a distance `d` such that the total work `W` done by the force is equal to the total kinetic
energy `E` of the car.

Think like this, our car has mechanical energy, and
this energy must be transformed into heat by the brakes. Transforming mechanical energy into
heat requires work, and this work is done by the force of friction. So the total work done by
the force must be equal to the total energy of our car, since we want to stop the car.

Now, the work done by a constant force `F`applied along a distance `d`is equal
to

`W = Fd`

But we have that `W = 392000 J` , by the
last calculation, and `d = 105 m`as given. Plugging in the values and solving for `F`gives the
value of the force:

`F = (W)/(d) = (392000 J)/(105 m) = (3733.33 N)`

Thus, the friction between the tires and the road must be approximately
`3733.33 N`. This is close to the force required to lift a `373 kg`weight!