60 *1000/3600 =16.7 m/s
average speed during stop = 8.33 m/s
time to stop t = 35/8.33 = 4.2 s
acceleration = - 16.7/ 4.2 = 3.98 m/s^2
m a = mu m g
mu = a/g
3.98 / g = .405
A 1250 kg car travelling at 60.0 km/h comes to a sudden stop in 35 m. What is the coefficient of friction acting on the brakes?
2 answers
Given
Vi=60km/h = (60/3.6) = 16.7m/s
Vf=0m/s
d=35m
a= (vf^2 - vi^2)/t = (0^2 - 16.7^2)/35
= -3.95m/s^2
Fnet=ma
=1250 x (-3.95)
=-4960.3
Fn=Fg=mg
=(1250kg)(9.8m/s^2)
=12250
Fapp= 0N
Ff =Fapp-Fnet
=0-(-4960.3)
=4960.3
mu =Ff/fn
=4960.3/12,250
mu =0.4
Vi=60km/h = (60/3.6) = 16.7m/s
Vf=0m/s
d=35m
a= (vf^2 - vi^2)/t = (0^2 - 16.7^2)/35
= -3.95m/s^2
Fnet=ma
=1250 x (-3.95)
=-4960.3
Fn=Fg=mg
=(1250kg)(9.8m/s^2)
=12250
Fapp= 0N
Ff =Fapp-Fnet
=0-(-4960.3)
=4960.3
mu =Ff/fn
=4960.3/12,250
mu =0.4
9 answers