A hockey puck of mass .2kg is sliding down an inclined plane with an angle of 15 degrees to the horizon. If the puck is moving with a constant speed, what is the coefficient of kinetic friction between the puck and the plane?
mgsinα=F(fr) = μN= μmgcosα
sinα= μcosα
μ=tanα
To find the coefficient of kinetic friction between the puck and the plane, we need to analyze the forces acting on the puck.
First, we need to determine the net force acting on the puck. Since the puck is moving with a constant speed, we know that the net force must be zero.
The forces acting on the puck are:
1. The force of gravity (mg), which acts straight downward and can be split into two components: one parallel to the inclined plane and one perpendicular to it.
2. The normal force (N), which acts perpendicular to the inclined plane.
3. The force of kinetic friction (f_k) parallel to the inclined plane.
The parallel component of the gravitational force can be found by multiplying the gravitational force (mg) by the sine of the angle between the inclined plane and the horizon:
F_parallel = mg * sin(angle)
The perpendicular component of the gravitational force can be found by multiplying the gravitational force (mg) by the cosine of the angle between the inclined plane and the horizon:
F_perpendicular = mg * cos(angle)
Since the puck is sliding down the inclined plane with a constant speed, the force of kinetic friction must be equal in magnitude and opposite in direction to the parallel component of the gravitational force:
f_k = F_parallel
Now, we can solve for the coefficient of kinetic friction using the formula:
f_k = μ * N
where μ is the coefficient of kinetic friction and N is the normal force. However, the normal force is equal in magnitude and opposite in direction to the perpendicular component of the gravitational force:
N = F_perpendicular
Thus, we can rewrite the equation as:
f_k = μ * F_perpendicular
By substituting the previously found expressions for f_k and F_perpendicular, we get:
F_parallel = μ * mg * cos(angle)
Since we know that F_parallel is equal to mg * sin(angle), we can equate the two equations:
mg * sin(angle) = μ * mg * cos(angle)
Now, we can cancel out the mass (mg) from both sides and solve for the coefficient of kinetic friction (μ):
sin(angle) = μ * cos(angle)
μ = sin(angle) / cos(angle)
μ = tan(angle)
Plugging in the value of the angle (15 degrees), we can calculate the coefficient of kinetic friction (μ):
μ = tan(15 degrees)
Using a calculator, we find that μ ≈ 0.267.
Therefore, the coefficient of kinetic friction between the puck and the plane is approximately 0.267.