Question

A 1370-kg car is skidding to a stop along a horizontal surface. The car decelerates from 27.6 m/s to a rest position in 3.15 seconds. Assuming negligible air resistance, determine the coefficient of friction between the car tires and the road surface.

Answer 1

[ F_f = \mu F_n\]

Normal Force
\[ F_n = mg = 1370 * 9.81 = 13439.7 N\]

Friction Force
\[ F_f = ma\]

car decelerates from 27.6 m/s to 0 in 3.15 sec

\[ a = \frac{dv}{dt} = \frac{27.6-0}{3.15-0} = 8.7619 \space \frac{m}{s^2} \]

therefor:

\[ F_f = 1370 * 8.7619 = 12003.8 N\]
and
\[ \mu = \frac{F_n}{F_f} = \frac{13439.7}{12003.8} = 1.1196\]

Answer 2

the answer is 0.894

Answer 3

\[F_{fr}=ma\]
\[\mu mg=ma\]
\[\mu g=\frac{ v }{ t }\]
\[\mu=\frac{ v }{g t }=0,894\]

Answer 4

the last part i did is wrong:

u = Ff/Fn = 12003.8/134349.7 = 0.8931

Answer 5

yes , friction force is proportional to normal force

Answer 6

thank you all!