A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches 30°, the box starts to slip, and it slides 2.4 m down the plank in 3.6 s at constant acceleration.
(a) What is the coefficient of static friction?
(b) What is the coefficient of kinetic friction?
a) tan 30
b) write the equation of motion, with a retarding force proportional to the kinetic friction coefficient, Uk. Compute the acceleration from the fact it takes 3.6 s to make it down the plank, starting at zero velocity. Solve for Uk
To determine the coefficients of static and kinetic friction, we can use the information given about the angle of inclination, distance traveled, and time taken.
(a) To find the coefficient of static friction, we first need to determine the acceleration of the box when it starts to slip. This can be done using the equations of motion. The equation for distance traveled can be written as:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity (which is zero in this case as the box starts from rest), a is the acceleration, and t is the time taken.
Given that s = 2.4 m and t = 3.6 s, we can rearrange the equation to solve for acceleration:
2.4 = 0 + (1/2)a(3.6)^2
Simplifying the equation:
2.4 = 1.8a
a = 2.4/1.8 = 1.33 m/s^2
Now, to find the coefficient of static friction:
μs = tan(θ)
where μs is the coefficient of static friction and θ is the angle of inclination.
Given that θ = 30°, we can substitute this value in the equation:
μs = tan(30°)
Using a calculator, we find:
μs ≈ 0.58
(b) To find the coefficient of kinetic friction, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity (which is zero in this case as the box starts from rest), a is the acceleration, and t is the time taken.
Given that s = 2.4 m, t = 3.6 s, and the acceleration is constant, we can rearrange the equation to solve for acceleration:
2.4 = 0 + (1/2)a(3.6)^2
Simplifying the equation:
2.4 = 1.8a
a = 2.4/1.8 = 1.33 m/s^2
The acceleration obtained here is the same as the one we found for the coefficient of static friction, indicating that the box slides down the plank with the same acceleration as when it starts to slip.
Now, to find the coefficient of kinetic friction:
μk = a/g
where μk is the coefficient of kinetic friction, a is the acceleration, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values:
μk = 1.33/9.8 ≈ 0.14
Therefore, the coefficient of static friction is approximately 0.58, and the coefficient of kinetic friction is approximately 0.14.