# Friction

• Nov 11th 2009, 01:18 PM
Caturdayz
Friction
A 2500 Kg car is heading east at 20 m/s. IT slams on the breaks to avoid hitting a deer. It takes the car 50 meters to stop, what is the Uk between the car and the road.
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Well I've been absent for school for two days, which is coincidentally when he was teaching this stuff.

So I'm not even exactly sure what Uk is.

I assume I'd use V^2=V0^2+2da ?
• Nov 11th 2009, 01:40 PM
BabyMilo
what is the Uk? the distance? the acceleration or deceleration?
so you want the friction? oh you want to find out muon not U. O.o
• Nov 11th 2009, 01:55 PM
BabyMilo
I assume you want Friction

yes, you use $v^2=u^2+2ax$

where,

* v=0m/s
* u=20m/s
* a=?
* x=50m

sub the numbers in

$0^2=20^2+2*a*50$

rearrange the equation to find a.

since Newton's second law defines $F=ma$ where F= force of overall system not friction

using the equation

$2500g-F=-2500a$ where F = friction

then rearrange the equation to find F.

you should get $a=-4ms^{-2}$
and $F=14500N$

where $g=9.8ms^{-2}$

To find out the reaction force, using the y directional force:
$-2500g+R=0$

Then $F=\mu*R$

rearrange to find $\mu$
• Nov 11th 2009, 01:58 PM
skeeter
Quote:

Originally Posted by Caturdayz
A 2500 Kg car is heading east at 20 m/s. IT slams on the breaks to avoid hitting a deer. It takes the car 50 meters to stop, what is the Uk between the car and the road.
__

Well I've been absent for school for two days, which is coincidentally when he was teaching this stuff.

So I'm not even exactly sure what Uk is.

I assume I'd use V^2=V0^2+2da ?

$\mu_k$ is the coefficient of kinetic friction.

$\mu_k = \frac{f_k}{N}$ , where $f_k$ is the force of sliding friction, and $N$ is the normal force between the surfaces in contact.

using the equation $v_f^2 = v_0^2 + 2a(\Delta x)$ , you can determine the car's acceleration ... then use Newton's 2nd Law of motion,
$F_{net} = ma$ , substituting $f_k$ for the net force ... then you can find $\mu_k$ from the equation above.