A skier of mass 64 kg starts from rest at the top of an 18° slope, points her skis straight down the slope and lets gravity pull her down the slope. The slope is 65 m long. Her speed at the bottom is 15 m/s. (a) Assuming zero air resistance, what is the magnitude of the frictional force between her skis and the snow? (b) What is the coefficient of kinetic friction between her skies and the snow?
(a) Subtract her final kinetic energy (MV^2/2) from her initial potential energy (M g H). That is the energy lost to friction. Divide that energy loss by the diatance she travels (65 m) to get the average friction force, f.
(b) f = muk * cos 18 * M * g
Solve for muk
thnks so much
To solve this problem, we can break it down into two parts.
Part (a): Finding the magnitude of the frictional force
We can use the concept of conservation of mechanical energy to find the frictional force. The initial mechanical energy is equal to the final mechanical energy, neglecting air resistance.
The initial mechanical energy is given by the gravitational potential energy at the top of the slope:
Initial mechanical energy = m * g * h,
where m is the mass of the skier, g is the acceleration due to gravity (9.8 m/s²), and h is the vertical height of the slope.
The final mechanical energy is given by the kinetic energy at the bottom of the slope:
Final mechanical energy = (1/2) * m * v²,
where v is the speed of the skier at the bottom of the slope.
Since there is no air resistance, we can equate the initial and final mechanical energies:
m * g * h = (1/2) * m * v².
Substituting the given values:
64 kg * 9.8 m/s² * h = (1/2) * 64 kg * (15 m/s)².
Simplifying the equation:
h = (15 m/s)² / (2 * 9.8 m/s²).
Calculating h:
h ≈ 11.6 m.
Now, we can find the magnitude of the frictional force at the bottom of the slope.
The frictional force can be calculated using the following equation:
Frictional force = M * g * μ,
where M is the mass of the skier, g is the acceleration due to gravity, and μ is the coefficient of kinetic friction between the skis and the snow.
Substituting the given values:
Frictional force = 64 kg * 9.8 m/s² * μ.
Since the frictional force is equal to the force required to accelerate the skier (assuming no air resistance) and we know the final speed, we can use the equation:
Frictional force = M * a,
where a is the acceleration of the skier.
Substituting the known values:
Frictional force = 64 kg * (15 m/s)² / 65 m.
Simplifying the equation:
Frictional force ≈ 23.43 kg m/s².
Therefore, the magnitude of the frictional force between the skis and the snow is approximately 23.43 N.
Part (b): Finding the coefficient of kinetic friction
To find the coefficient of kinetic friction, we can use the equation:
μ = (Frictional force) / (M * g),
where M is the mass of the skier and g is the acceleration due to gravity.
Substituting the known values:
μ = 23.43 N / (64 kg * 9.8 m/s²).
Calculating μ:
μ ≈ 0.037.
Therefore, the coefficient of kinetic friction between the skis and the snow is approximately 0.037.
To find the magnitude of the frictional force between the skier's skis and the snow, we can use the principles of Newton's laws of motion and the concept of conservation of energy.
(a) The first step is to calculate the gravitational potential energy at the top of the slope and the kinetic energy at the bottom. The change in potential energy will be equal to the change in kinetic energy, as there is no other energy loss or gain.
The gravitational potential energy (PE) is given by the formula:
PE = m * g * h
where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height change.
In this case, the height change is given by:
h = length of the slope * sin(angle of the slope)
h = 65 m * sin(18°)
The change in potential energy is then:
∆PE = m * g * h
The kinetic energy (KE) at the bottom of the slope is given by:
KE = 1/2 * m * v^2
where v is the speed at the bottom of the slope (15 m/s).
Since energy is conserved, we can equate the change in potential energy to the change in kinetic energy:
∆PE = ∆KE
Therefore:
m * g * h = 1/2 * m * v^2
Simplifying the equation and solving for the gravitational force (m * g):
m * g = (1/2 * m * v^2) / h
Now, we can find the magnitude of the gravitational force acting down the slope:
m * g = (1/2 * 64 kg * (15 m/s)^2) / (65 m * sin(18°))
Calculating this, we get:
m * g ≈ 115.42 N
Since the gravitational force is acting down the slope, the magnitude of the frictional force will be equal to the gravitational force (due to the skier maintaining a constant speed).
Therefore, the magnitude of the frictional force between the skier's skis and the snow is approximately 115.42 N (using the given mass of the skier).
(b) To find the coefficient of kinetic friction, we can use the formula:
frictional force = coefficient of kinetic friction * normal force
In this case, the magnitude of the frictional force is equal to the gravitational force (115.42 N).
Since the normal force is equal to the component of the gravitational force perpendicular to the slope, it can be calculated as:
normal force = m * g * cos(angle of the slope)
normal force = 64 kg * 9.8 m/s^2 * cos(18°)
Calculating this, we get:
normal force ≈ 609.17 N
Substituting the values into the formula for the frictional force, we have:
115.42 N = coefficient of kinetic friction * 609.17 N
Solving for the coefficient of kinetic friction, we find:
coefficient of kinetic friction ≈ 0.189
Therefore, the coefficient of kinetic friction between the skier's skis and the snow is approximately 0.189 (using the given values and calculations).