Asked by Million Tefera
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
Answered by
Anonymous
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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