AP® Physics B: Newtonian Mechanics Drill 2, Problem 3. What is the coefficient of friction between the biscuit and the court?
|AP Physics B||Newtonian Mechanics|
|AP Physics B/C||Newtonian Mechanics|
|AP Physics C||Newtonian Mechanics|
With a push of the broom, a puck... which is called a biscuit,
and no, we're not making that up... goes flying down the lane, hopefully coming to a stop
at the perfect spot.
It's like curling for people without convenient access to ice.
If an elderly player is able to give the 4 kilogram biscuit an initial speed of 5 meters
per second, and it comes to rest after 10 meters, what
is the coefficient of friction between the biscuit and the court?
And here are the possible answers...
This problem is a two-parter.
We first have to find how fast the biscuit is decelerating,
then find the coefficient of friction.
Keep in mind that there is no correlation between this and how fast
your biscuit decelerates after you send it sliding down your esophagus.
First, we know that the biscuit is moving at 5 meters per second initially,
and decelerates to a stop in 10 meters.
Not a lot of information, but it's enough.
We can use v squared = v initial squared plus 2 a x, as our core formula.
Plugging in what we know, we get 0 squared is equal to... 5 squared plus 2a times 10.
With some simple algebra, we get -25 is equal to 20a,
so a is equal to -5/4 meters per second squared.
Now that we have our acceleration, this problem is easy.
F = ma, and then F = u times the normal force.
First, to find out how much force friction is applying to the biscuit, we multiply the
mass of the biscuit by the acceleration, which we found in the last part.
4 kilograms times -5/4 meters per second is equal to 5 Newtons of force.
Sure... go ahead and chase that biscuit with a few Fig Newtons...
Finally, the force of friction is equal to mu times the normal force.
The force of friction is equal to -5 Newtons, so we plug that into the equation.
We're missing the normal force, but since the biscuit is on a flat plane,
that's just equal to mass times acceleration due to gravity, or 4(-10).
Solving for mu, we get that the coefficient of friction is equal to 1/8...
...or choice (A).
Now to deal with the friction in Grandma and Grandpa's relationship...