A 10-g fridge magnet is supposed to hold a 2.0-g paper on the stainless steel fridge door. The normal force of 0.19 N is insufficient to keep them in place, and the paper and magnet accelerate down the fridge at 0.150 m/s2 when the fridge door is slammed. What is the coefficient of kinetic friction between the steel door and the paper�
To find the coefficient of kinetic friction between the steel door and the paper, we can use the equation for the net force acting on the paper:
Net force = frictional force - normal force
The net force can also be calculated using the equation:
Net force = mass × acceleration
Given:
Mass of the fridge magnet (m1) = 10 g = 0.01 kg
Mass of the paper (m2) = 2.0 g = 0.002 kg
Acceleration (a) = 0.150 m/s²
Normal force (Fn) = 0.19 N
First, let's calculate the net force acting on the system (paper and magnet):
Net force = (m1 + m2) × a
= (0.01 kg + 0.002 kg) × 0.150 m/s²
Next, let's find the frictional force:
Frictional force = Net force - Normal force
We know that the frictional force also equals the coefficient of kinetic friction (μ) multiplied by the normal force:
Frictional force = μ × Fn
Therefore,
μ × Fn = Net force - Normal force
Now we can solve for the coefficient of kinetic friction (μ):
μ = ((m1 + m2) × a - Fn) / Fn
Substituting the given values:
μ = ((0.01 kg + 0.002 kg) × 0.150 m/s² - 0.19 N) / 0.19 N
Calculating:
μ = (0.016 kg × 0.150 m/s² - 0.19 N) / 0.19 N
= (0.0024 kg·m/s² - 0.19 N) / 0.19 N
= 0.0024/0.19 - 1
≈ 0.0126 - 1
≈ -0.9874
The coefficient of kinetic friction between the steel door and the paper is approximately -0.9874.
To find the coefficient of kinetic friction between the steel door and the paper, we can use the equation:
μk = (m * a) / (m * g)
where
μk = coefficient of kinetic friction,
m = mass of the paper,
a = acceleration of the paper,
g = acceleration due to gravity.
Given values:
Mass of the paper, m = 2.0 g = 0.002 kg
Acceleration of the paper, a = 0.150 m/s^2
Acceleration due to gravity, g = 9.8 m/s^2
Plugging in the values:
μk = (0.002 kg * 0.150 m/s^2) / (0.002 kg * 9.8 m/s^2)
μk = (0.0003 N) / (0.0196 N)
μk = 0.0153
Therefore, the coefficient of kinetic friction between the steel door and the paper is approximately 0.0153.