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.

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