Sukey is riding her bicycle when she sees a fallen tree 42 m ahead. If the coefficient of friction is .36 and she is travelling at 50 km/ hr and she and the bike have a mass of 95 kg, how much stopping distance does she need. I got as far as her speed equals 13.9 m/sec. Do we need her Normal force (95 x -9.8m/sec 2) = 930N.
10c-<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C-<0.09>=f(1.33)
d=?
Sorry, I don't understand all the abbreviations.
Sorry, I don't understand all the abbreviations.
I figure that her stopping speed, 0 = 13.9m/sec + 2ad where a = F/m but here I get stuck. I don't know exactly what to use for F
27.3m
To calculate the stopping distance, we need to consider the forces acting on Sukey and her bicycle.
First, let's convert Sukey's speed from km/hr to m/sec. You correctly calculated that her speed is 13.9 m/sec.
Now we need to calculate the force of friction (Ffr) between the bicycle and the ground. The formula for friction is:
Ffr = coefficient of friction * normal force
Given that the coefficient of friction is 0.36 and the normal force can be calculated as mass * acceleration due to gravity (Fnormal = mass * g), we can substitute the values and find:
Ffr = 0.36 * (95 kg * 9.8 m/sec^2) = 330.48 N
Next, let's calculate the deceleration (a) using the equation:
force = mass * acceleration
Since the force in this case is the force of friction (Ffr) and the mass is the total mass of Sukey and her bicycle (95 kg), we have:
Ffr = mass * acceleration
330.48 N = 95 kg * acceleration
acceleration = 330.48 N / 95 kg = 3.47 m/sec^2 (rounded to two decimal places)
Now, we can use the basic kinematic equation to calculate the stopping distance (d):
v^2 = v0^2 + 2a * d
where:
v = final velocity (which is 0 m/sec as Sukey needs to stop)
v0 = initial velocity (13.9 m/sec)
a = acceleration (-3.47 m/sec^2)
d = stopping distance (unknown)
Substituting the values, we get:
0 = (13.9 m/sec)^2 + 2 * (-3.47 m/sec^2) * d
0 = 193.21 m^2/sec^2 - 6.94 m/sec^2 * d
6.94 m/sec^2 * d = 193.21 m^2/sec^2
d = 193.21 m^2/sec^2 / 6.94 m/sec^2 = 27.82 m (rounded to two decimal places)
Therefore, Sukey needs a stopping distance of approximately 27.82 meters.