# Thread: Friction question.

1. ## Friction question.

In a game. A coin of mass 5 grams is struck and slides across a wooden board.
If the co-efficient of friction of 0.4 and the coin is struck with a speed of 1.5m/s.
How far will it slide until it comes to rest?

I got this far... And now i'm stuck...(image attached)

Limiting friction = 0.05g * 0.4 = 0196N.
D - 0.196N = 0.05a

Well i have that u = 1.5m/s and v = 0...however i need acceleration in order to use any of the equations to find the distance..and I'm not sure how to do it...

Thank-you

2. The only force acting on the coin after pushing it is the force due to friction, so:
$F_{net} = F_{fr}$
$\Rightarrow m\vec{a} = \mu \: mg$ (Since $F_{fr} = \mu F_{n} = \mu F_{g}$ where $F_{n}$ is the normal force balanced by the force due to gravity $F_{g}$)

Now that you've found acceleration, apply the kinematics equation:
$\vec{v_{f}} = \vec{v_{i}} + \vec{a}t$

where $v_{f}$ is your final velocity (which is 0 if it's coming to rest) and $v_{i}$ is your initial velocity.

3. Originally Posted by o_O

Now that you've found acceleration, apply the kinematics equation:
$\vec{v_{f}} = \vec{v_{i}} + \vec{a}t$
I'm getting something thats /100 the answer. (answers 28.7 but im getting 0.287)
$a = \frac{-0.196}{0.05}$

$a = -3.92$

First i did
v = 0. u = 1.5m/s
$v^2 - u^2 = 2as
-2.25 = 2 * -3.92 * s
$

s = 0.287

So i tried with v = u + at
$0 = 1.5 = 3.92t
t = 0.38
$

When i sub t, v and u into
$s = 0.5(v + u)t$

I also get s = 0.287.. :S

Thanks

4. Originally Posted by AshleyT
I'm getting something thats /100 the answer. (answers 28.7 but im getting 0.287)
$a = \frac{-0.196}{0.05}$

$a = -3.92$

First i did
v = 0. u = 1.5m/s
$v^2 - u^2 = 2as
-2.25 = 2 * -3.92 * s
$

s = 0.287

So i tried with v = u + at
$0 = 1.5 = 3.92t
t = 0.38
$

When i sub t, v and u into
$s = 0.5(v + u)t$

I also get s = 0.287.. :S

Thanks
Do you pay attention to units .....?

Your answer is in metres. No doubt the books answer is in centimetres.