Burning rubber

Classical Mechanics Level 2

You're quietly sitting by a street one day, enjoying the sun, when you notice a speeding car traveling at 20 m/s down the street. The driver notices that there's a stop sign in the street and hits the brakes 40 m before the stop sign, coming to a stop right at the sign itself. What is the coefficient of friction between the car's tires and the road?

Details and assumptions

Assume that once the driver hits the brakes the wheels no longer turn but instead slide against the road.

The answer is 0.51.

7 solutions

Nahom Yemane
Jan 3, 2014

LOSS of energy: $\frac{1}{2}mv^2=\frac{1}{2}m(20^2)=200m$ where $m$ is mass

Work done by friction: $\mu Rd$ where $R$ and $d$ are Normal Reaction force and distance travelled respectively. $R=mg$ so no vertical accerleration By Work-Energy principle we have: $\mu Rd= 200m$

so $\mu mgd= 200m$

$d=40$

$40\mu g= 200$

$\mu = \frac{200}{40g}= \boxed{0.501} (3s.f$ )

Aravind Raj
Mar 4, 2014

Stoping Distance S=u^2/2a a= µg 40=400/2 µ*9.8 µ=0.5102

Abhinav Mishra
Aug 21, 2015

Using Newton's equation of motions, v^2 = u^2 + 2as where s is the displacement. Given : v = 0, u = 20m/s, s = 40m. Plugging these values in the above equation we can find out the deceleration of the car. (a = -5m/s^2) Using Newton's Second Law, F=ma and Using Laws of friction, F=μN where N is the normal reaction. N=W=mg (weight of the body) Equation these, we get, ma = μmg. Plugging the values of a and g, μ can be found as (-5m/s^2)/(-9.81m/s^2) = 0.5096 = 0.51.