A skier starts from rest at the top of a frictionless incline of height 20.0 m, as shown in the figure. At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.208. Neglect air resistance.Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.208. Assume that θ = 20.0°.
theta= 20
Skiers speed at bottom= 19.8 m/s
m=96.1
x= 96.1/(tan(20))
To find the horizontal distance the skier travels before coming to rest, we can use the equations of motion.
First, let's find the speed of the skier at the bottom of the incline using conservation of mechanical energy. The potential energy at the top of the incline will be converted into kinetic energy at the bottom.
Since the incline is frictionless, the only form of energy that changes is the potential energy:
m * g * h = (1/2) * m * v^2
Here,
m = mass of the skier = 96.1 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the incline = 20.0 m
v = speed of the skier at the bottom
Therefore,
v = sqrt(2 * g * h)
Plugging in the values:
v = sqrt(2 * 9.8 * 20.0) = 19.8 m/s
Now, let's find the horizontal distance the skier travels on the horizontal surface while experiencing friction.
The formula for calculating the distance covered with uniform acceleration is:
x = (v^2 - u^2) / (2 * a)
Here,
v = final velocity of the skier = 0 m/s (since they come to rest)
u = initial velocity of the skier = 19.8 m/s
a = acceleration due to friction = -μ * g
Given that the coefficient of kinetic friction is 0.208, we have:
a = -0.208 * 9.8 = -2.0384 m/s^2
Plugging in the values:
x = (0^2 - 19.8^2) / (2 * -2.0384)
x = -396.04 / -4.0768
x = 96.949 m
Since distance cannot be negative, we take the magnitude of the result:
x = |96.949| = 96.949 m
Therefore, the skier travels approximately 96.949 meters horizontally before coming to rest.