A 64.5-kg skier moving horizontally at 4.83 m/s encounters a 16.7° incline.
a) How far up the incline will the skier move before she momentarily stops, ignoring friction?
b) How far up the incline will the skier move if the coefficient of kinetic friction between the skies and snow is 0.103?
a) v^2/(2g) = h. h/sin(theta) = d
b) v^2/(2*g*mu) = h. h/(mu*sin(theta)) = d
To solve these problems, we will use Newton's laws of motion and the concept of work and energy.
a) To find how far the skier moves up the incline before momentarily stopping, we need to consider the forces acting on the skier. Since we are ignoring friction, the only force acting on the skier is the component of the gravitational force pulling her down the incline. This force is given by:
F = mg sin(θ)
Where:
F is the force acting on the skier (N),
m is the mass of the skier (kg),
g is the acceleration due to gravity (9.8 m/s^2), and
θ is the angle of the incline (16.7°).
The gravitational force is acting in the opposite direction to the skier's motion. When the skier momentarily stops, the net force acting on her is zero. So we have:
F = ma
Where:
a is the acceleration of the skier (m/s^2).
Since the mass is constant, we can cancel it from both equations:
mg sin(θ) = ma
Now, we need to find the acceleration. In this case, the acceleration is the component of gravity along the incline, which is given by:
a = g sin(θ)
Substituting this value into the equation, we get:
mg sin(θ) = mg sin(θ)
Since both sides of the equation are equal, the motion along the incline will stop momentarily, regardless of the mass of the skier. Therefore, it doesn't matter whether the skier weighs 64.5 kg or any other mass. To find how far the skier moves up the incline before stopping, we need to use the equation for displacement along an inclined plane:
d = (v^2 - u^2) / (2g sin(θ))
Where:
d is the displacement along the incline (m),
v is the final velocity of the skier (m/s), and
u is the initial velocity of the skier (m/s) (given as 4.83 m/s).
Since the skier stops (final velocity v = 0 m/s), we have:
d = (0^2 - 4.83^2) / (2 * 9.8 * sin(16.7°))
Using a calculator:
d ≈ -12.02 m
The negative sign indicates that the displacement is in the opposite direction to the initial motion. Therefore, the skier will move approximately 12.02 meters down the incline before stopping momentarily.
b) Now let's consider the case when the coefficient of kinetic friction between the skis and snow is 0.103. In this case, we need to account for the frictional force acting on the skier.
The frictional force can be calculated using the equation:
F_friction = μ * m * g * cos(θ)
Where:
F_friction is the frictional force (N),
μ is the coefficient of kinetic friction,
m is the mass of the skier (kg),
g is the acceleration due to gravity (9.8 m/s^2), and
θ is the angle of the incline (16.7°).
The frictional force acts in the opposite direction to the motion, so we can write:
F_friction = -μ * m * g * cos(θ)
Now, the net force acting on the skier can be calculated as:
F_net = mg sin(θ) + F_friction
Since the skier moves uphill, the net force is acting in the positive direction, so we have:
F_net = ma
Now, let's substitute the values into the equation:
mg sin(θ) + (-μ * m * g * cos(θ)) = ma
Since the mass is constant, we can again cancel it from both sides of the equation:
g sin(θ) - μ * g * cos(θ) = a
Now, we need to substitute the value of acceleration into the equation for displacement:
d = (v^2 - u^2) / (2(g sin(θ) - μ * g * cos(θ)))
Using the given values:
u = 4.83 m/s,
v = 0 m/s,
θ = 16.7°,
μ = 0.103,
g = 9.8 m/s^2,
We can now substitute these values into the equation and calculate the displacement d:
d = (0^2 - 4.83^2) / (2(9.8 sin(16.7°) - 0.103 * 9.8 cos(16.7°)))
Using a calculator, we find:
d ≈ 6.49 m
Therefore, with a coefficient of kinetic friction of 0.103, the skier will move approximately 6.49 meters up the incline before coming to a stop.