friction force = -0.1 m g
F = - 0.1 m g = m a
so acceleration = a = -0.1 g
v = Vi + a t
if v = 0
a t = -Vi
if Vi =15
-0.1 g t = -15
t = 150/g
x = Vi t + (1/2) a t^2
x = 15 (150/g) + (1/2)(-.1g)(150^2) / g^2
= 15 (150/g) - 75 * 150 (.1) / g
= 15(150/g) - 7.5 (150/g) = 7.5 (150/g)
now do it using 30 instead of 15 m/s
In a car the brakes stop while friction btn the road surface stop the car .On a wet road the coefficient of kinetic friction between the road surface and the tyres is 0.1.Two cars A and B are travelling at a speed of 15m/s and 30m/s respectively .Brakes are suddenly applied on each of the cars .How far will the cars travel before coming to rest
2 answers
by the way if you know the energy methods you can say the work done by friction = change of kinetic energy
0.1 m g * x = (1/2) m v^2
x = 5 v^2/g
if v = 15
x = 5*225/g which is 1125/g which is the same as 7.5(150/g)
0.1 m g * x = (1/2) m v^2
x = 5 v^2/g
if v = 15
x = 5*225/g which is 1125/g which is the same as 7.5(150/g)