A 2.0-kg box rests on a plank that is inclined at an angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 400 N/m. If the coefficient of static friction between the box and the plank is 0.22, what is the maximum amount the spring can be stretched and the box remain at rest?

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To find the maximum amount the spring can be stretched and the box remain at rest, we need to consider the forces acting on the box.

Let's break down the forces acting on the box along the inclined plank:

1. Weight force (mg): The weight force acts vertically downward and is given by mg, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, mg = 2.0 kg * 9.8 m/s^2 = 19.6 N.

2. Normal force (N): The normal force acts perpendicular to the inclined plank, preventing the box from sinking into the plank.
The normal force can be calculated using the formula N = mg * cosθ, where θ is the angle of inclination.
In this case, θ = 65 degrees, so cosθ = cos(65) ≈ 0.4226.
Therefore, N ≈ 19.6 N * 0.4226 ≈ 8.28 N.

3. Friction force (f): The static friction force acts parallel to the inclined plane, opposing the motion of the box.
The maximum static friction force can be calculated using the formula f = μs * N, where μs is the coefficient of static friction.
In this case, μs = 0.22 and N = 8.28 N, so f ≈ 0.22 * 8.28 N ≈ 1.822 N.

Since the box is at rest, the component of the weight force along the inclined plane is balanced by the static friction force.
Therefore, mg * sinθ = f.

Now let's calculate the maximum amount the spring can be stretched:

The restoring force of the spring (Fs) is given by Fs = -k * x, where k is the force constant of the spring and x is the stretching or compression of the spring.
In this case, k = 400 N/m.

At maximum stretching, the restoring force of the spring should be equal to the static friction force acting on the box.
Therefore, -k * x = f.

Substituting the values, we have:

-400 N/m * x = 1.822 N.

Solving for x, we find:

x ≈ 1.822 N / (-400 N/m) ≈ -0.0046 m.

Since stretching is a positive distance, we take the absolute value:

|x| ≈ 0.0046 m.

Hence, the maximum amount the spring can be stretched while the box remains at rest is approximately 0.0046 meters.