An Olympic skier moving at 19 m/s down a 26 degree slope encounters a region of wet snow and slides 136 m before coming to a halt. What is the coefficient of friction between the skis and the snow? ( g = 9.81 m/s2)
.64
0.638 is correct
To find the coefficient of friction between the skis and the snow, we need to consider the forces acting on the skier and use the concepts of physics related to motion.
First, let's break down the forces acting on the skier:
1. The gravitational force (mg), where m is the mass of the skier and g is the acceleration due to gravity.
2. The force of friction (Ff) between the skis and the snow.
The gravitational force can be divided into two components: the component parallel to the slope and the component perpendicular to the slope.
The component parallel to the slope is given by mg * sin(theta), where theta is the angle of the slope (26 degrees in this case).
The frictional force can be further divided into two components: the component parallel to the slope and the component perpendicular to the slope. Since the skier comes to a halt, the component parallel to the slope (Ff_par) will be opposing the component parallel to the slope caused by gravity.
Now, we can set up an equation of motion, using Newton's second law:
Fnet = ma
Since the skier is coming to a halt, the net force (Fnet) acting on the skier is equal to zero.
Thus, we have:
Ff_par - mg * sin(theta) = 0
Rearranging the equation, we get:
Ff_par = mg * sin(theta)
Now, we need to express Ff_par (the frictional force parallel to the slope) in terms of the coefficient of friction (μ) and the normal force (Fn) acting on the skier. The normal force can be calculated as the component of gravity perpendicular to the slope:
Fn = mg * cos(theta)
The frictional force can be expressed as:
Ff_par = μ * Fn
Substituting the expressions for Ff_par and Fn:
μ * mg * cos(theta) = mg * sin(theta)
The mass (m), acceleration due to gravity (g), and angle of the slope (theta) are given in the problem. We can substitute these values to find the coefficient of friction (μ):
μ = (g * sin(theta)) / (g * cos(theta))
μ = tan(theta)
Let's calculate the coefficient of friction:
μ = tan(26 degrees)
Using a scientific calculator, we find:
μ ≈ 0.4877
Therefore, the coefficient of friction between the skis and the snow is approximately 0.4877.