Question
A car with mass of 1000kg travelling at speed of 108km/hr along level road is brought to stop uniformly in distance of 70m. What was the coefficient of friction between the tires and the the roads?
Answers
108 km/hr = 108,000 meters / 3600 seconds = 30 meters/seconds
d =70 = average speed * t = 15 t
so t = 70/15 = 14/3
d = Vi t + (1/2) a t^2
70 = 30 (14/3) + (1/2) a (14/3)^2
70 = 140+ 10.9 a
10.9 a = -70
a = - 6.43 m/s^2
F = m a = - 6430 Newtons
6430 = mu m g
so
mu = 6.430 / 9.81 = 0.655
d =70 = average speed * t = 15 t
so t = 70/15 = 14/3
d = Vi t + (1/2) a t^2
70 = 30 (14/3) + (1/2) a (14/3)^2
70 = 140+ 10.9 a
10.9 a = -70
a = - 6.43 m/s^2
F = m a = - 6430 Newtons
6430 = mu m g
so
mu = 6.430 / 9.81 = 0.655
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