A bicycle and rider of total weight 815 N travel at a speed of 40.0 km/h.
The rider hits the breaks and the bicycle slides to a full stop in a distance of 8.5 m.
Ignore air resistance and use g=9.8 m/s2.
What is the coefficient of kinetic friction between the tires (rubber) and the road (pavement)?
N = total normal reaction = 815N
m = N/g = 83.16 kg
v0 = initial velocity = 40 km/h = 11.1 m/s
v1 = final velocity = 0
S = braking distance = 8.5 m
a=(v1²-v0²)/2S
μN = ma
Solve for μ
Thank you. I just wanted to confirm that this was the right way to do the question.
Just another clarification. In the FBD for this question, would the kinetic friction force be in the positive horizontal direction i.e. in the direction the rider was moving before he braked?
Since kinetic frictional force always opposes motion, by definition, it is opposite in direction to the movement of the bicycle. Also, this is how the bicycle is decelerated.
To find the coefficient of kinetic friction between the tires (rubber) and the road (pavement), we can use the following equation:
Frictional force = coefficient of kinetic friction × normal force
First, let's find the normal force acting on the bicycle and rider. The normal force is the force exerted by the road on the bicycle. In this case, the normal force is equal to the weight of the bicycle and rider, which is 815 N.
Next, we need to find the frictional force. The frictional force is the force opposing the motion of the bicycle and rider, and it can be calculated using Newton's second law:
Frictional force = mass × acceleration
We know the mass of the system is equal to the weight divided by the acceleration due to gravity. So,
mass = weight / acceleration due to gravity
mass = 815 N / 9.8 m/s^2
Now, we need to find the acceleration. Since the bicycle and rider come to a full stop, their final velocity is 0. We can use the following kinematic equation to find the acceleration:
vf^2 = vi^2 + 2as
where
vf = final velocity (0 m/s, since the bicycle stops)
vi = initial velocity (40 km/h, which we need to convert to m/s)
s = distance (8.5 m)
a = acceleration
Converting the initial velocity to m/s:
vi = 40.0 km/h × (1000 m/1 km) / (3600 s/1 h)
vi = 11.1 m/s
Plugging in the values:
(0 m/s)^2 = (11.1 m/s)^2 + 2a(8.5 m)
Simplifying the equation:
0 m^2/s^2 = 123.21 m^2/s^2 + 17a
Rearranging the equation:
17a = -123.21 m^2/s^2
a = -123.21 m^2/s^2 / 17
a ≈ -7.25 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is consistent with the bicycle coming to a stop.
Now, we can calculate the frictional force:
Frictional force = mass × acceleration
Frictional force = (815 N / 9.8 m/s^2) × (-7.25 m/s^2)
Finally, we can use the frictional force equation mentioned earlier to find the coefficient of kinetic friction:
Frictional force = coefficient of kinetic friction × normal force
Rearranging the equation, we can solve for the coefficient of kinetic friction:
coefficient of kinetic friction = Frictional force / normal force
Plug in the values:
coefficient of kinetic friction = (Frictional force) / (815 N)
Now, calculate the frictional force and substitute it into the equation:
coefficient of kinetic friction = ((815 N / 9.8 m/s^2) × (-7.25 m/s^2)) / 815 N
After calculating, the coefficient of kinetic friction between the tires (rubber) and the road (pavement) is approximately -0.58.
Note: The negative sign may seem unusual, but it simply indicates the direction of the frictional force relative to the direction of motion. The magnitude of the coefficient of kinetic friction is what really matters in this context.