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AP Physics B/C Videos 8 videos

AP Physics B 2.1 Newtonian Mechanics
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AP Physics B 1.3 Newtonian Mechanics
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AP® Physics B: Newtonian Mechanics Drill 1, Problem 3. With what acceleration does lunch arrive?

AP Physics C 1.1 Newtonian Mechanics
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AP Physics C: Newtonian Mechanics Drill 1, Problem 1. Which of Newton's equations of motion would have to be modified to account for the airplane's...

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AP Physics B 2.3 Newtonian Mechanics 178 Views


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Description:

AP® Physics B: Newtonian Mechanics Drill 2, Problem 3. What is the coefficient of friction between the biscuit and the court?

Language:
English Language

Transcript

00:03

It's time for your daily dose of Shmoop... ...so open wide.

00:07

Shuffleboard, if you haven't heard of it, is an "old people" sport that retirees play

00:11

on cruise ships.

00:13

Grandma or Grandpa stands at one end of a narrow lane, and a grid of spaces with various

00:18

scores is at the other end.

00:21

With a push of the broom, a puck... which is called a biscuit,

00:25

and no, we're not making that up... goes flying down the lane, hopefully coming to a stop

00:29

at the perfect spot.

00:32

It's like curling for people without convenient access to ice.

00:36

If an elderly player is able to give the 4 kilogram biscuit an initial speed of 5 meters

00:41

per second, and it comes to rest after 10 meters, what

00:45

is the coefficient of friction between the biscuit and the court?

00:49

And here are the possible answers...

00:54

This problem is a two-parter.

00:57

We first have to find how fast the biscuit is decelerating,

01:02

then find the coefficient of friction.

01:08

Keep in mind that there is no correlation between this and how fast

01:12

your biscuit decelerates after you send it sliding down your esophagus.

01:16

First, we know that the biscuit is moving at 5 meters per second initially,

01:19

and decelerates to a stop in 10 meters.

01:22

Not a lot of information, but it's enough.

01:24

We can use v squared = v initial squared plus 2 a x, as our core formula.

01:34

Plugging in what we know, we get 0 squared is equal to... 5 squared plus 2a times 10.

01:44

With some simple algebra, we get -25 is equal to 20a,

01:48

so a is equal to -5/4 meters per second squared.

01:55

Now that we have our acceleration, this problem is easy.

01:59

F = ma, and then F = u times the normal force.

02:06

First, to find out how much force friction is applying to the biscuit, we multiply the

02:10

mass of the biscuit by the acceleration, which we found in the last part.

02:14

4 kilograms times -5/4 meters per second is equal to 5 Newtons of force.

02:20

Sure... go ahead and chase that biscuit with a few Fig Newtons...

02:24

Finally, the force of friction is equal to mu times the normal force.

02:29

The force of friction is equal to -5 Newtons, so we plug that into the equation.

02:34

We're missing the normal force, but since the biscuit is on a flat plane,

02:38

that's just equal to mass times acceleration due to gravity, or 4(-10).

02:45

Solving for mu, we get that the coefficient of friction is equal to 1/8...

02:51

...or choice (A).

02:52

Now to deal with the friction in Grandma and Grandpa's relationship...

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