A 2.0 kg box rests on a plank that is inclined at an angle of 65° above the horizontal. The upper end of the box is attached to a spring with a force constant of 17 N/m, as shown in the figure. If the coefficient of static friction between the box and the plank is 0.30, what is the maximum amount the spring can be stretched and the box remain at rest?

Well, well, well, let's do some springy calculations, shall we? First things first, we gotta find the gravitational force acting on that box. So, we'll use the formula F = mg, where m is the mass of the box (2.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).

F = (2.0 kg) * (9.8 m/s²)
F = 19.6 N

Next up, we need to find the maximum static frictional force that can be exerted on the box. To do that, we use the formula Ffriction = μs * FN, where μs is the coefficient of static friction (0.30) and FN is the normal force acting on the box.

The normal force is the force perpendicular to the inclined plane, which can be found using the formula FN = m * g * cos(θ), where θ is the angle of inclination (65°).

FN = (2.0 kg) * (9.8 m/s²) * cos(65°)
FN ≈ 37.6 N

Now, let's calculate the maximum static frictional force.

Ffriction = (0.30) * (37.6 N)
Ffriction ≈ 11.3 N

The maximum amount the spring can be stretched is equal to the force acting on it, which is the sum of the gravitational force and the maximum static frictional force.

Max Stretch = F + Ffriction
Max Stretch = 19.6 N + 11.3 N
Max Stretch ≈ 30.9 N

So, the maximum amount the spring can be stretched while keeping the box at rest is approximately 30.9 N. And remember, don't stretch yourself too thin while solving physics problems!

To find the maximum amount the spring can be stretched before the box starts moving, we need to consider the forces acting on the box:

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

2. Normal force: This force acts perpendicular to the inclined plane and balances the component of the weight force that acts parallel to the inclined plane (N = mg cos θ). Here, θ is the angle of inclination (65°).

3. Static friction: This force acts parallel to the inclined plane and prevents the box from sliding down. The maximum static friction force is given by fs_max = μs * N, where μs is the coefficient of static friction (0.30 in this case).

4. Spring force: This force acts opposite to the displacement of the spring and is given by Hooke's law: F = k * x, where k is the force constant of the spring (17 N/m) and x is the displacement of the spring from its equilibrium position.

Now, we can set up an equation to find the maximum displacement of the spring:

fs_max = F (at the threshold of motion)
μs * N = k *x

To find N, we need to consider the force components acting on the box:

N = mg cos θ
N = 2.0 kg * 9.8 m/s^2 * cos(65°)
N = 2.0 kg * 9.8 m/s^2 * (0.4226) ≈ 8.215 N

Now, we can plug in the values into the equation for maximum static friction:

0.30 * 8.215 N = 17 N/m * x

Simplifying the equation:

2.4645 N = 17 N/m * x

Finally, we can solve for x:

x = 2.4645 N / (17 N/m)
x ≈ 0.145 meters (or 14.5 cm)

Therefore, the maximum amount the spring can be stretched before the box starts moving is approximately 0.145 meters or 14.5 cm.

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

Let's identify the forces acting on the box:

1. Weight (W): The force exerted by gravity on the box. It acts vertically downward and its magnitude is given by W = m * g, where m is the mass of the box and g is the acceleration due to gravity.

2. Normal force (N): The force exerted by the plank on the box perpendicular to the plank's surface.

3. Static friction force (Ff): The force opposing the motion between the box and the plank as long as they are at rest relative to each other. The magnitude of static friction can be found using the equation Ff = μs * N, where μs is the coefficient of static friction.

4. Spring force (Fs): The force exerted by the spring on the box. The magnitude of the force is given by Hooke's Law: Fs = k * Δx, where k is the force constant of the spring and Δx is the displacement of the spring from its equilibrium position.

Let's break down the weight (W) force into its components:

- W_parallel: The component of the weight parallel to the plank.
- W_perpendicular: The component of the weight perpendicular to the plank.

W_parallel = W * sinθ, where θ is the angle of inclination.

Now, let's set up the equilibrium conditions:

1. The sum of the forces in the horizontal direction must be zero since the box is at rest. The only horizontal force acting on the box is the static friction force (Ff).
∑F_horizontal = Ff - W_parallel = 0
Ff = W_parallel

2. The sum of the forces in the vertical direction must also be zero since the box is at rest. The forces in the vertical direction are the normal force (N) and the perpendicular component of weight (W_perpendicular).
∑F_vertical = N - W_perpendicular = 0
N = W_perpendicular

3. The maximum amount the spring can be stretched is when the force exerted by the spring (Fs) is equal to the static friction force (Ff). This occurs when the box is at the verge of sliding and just about to move.
Fs = Ff
k * Δx = μs * N

Now we can substitute the values given in the problem:

m = 2.0 kg (mass of the box)
θ = 65° (angle of inclination)
g = 9.8 m/s^2 (acceleration due to gravity)
k = 17 N/m (force constant of the spring)
μs = 0.30 (coefficient of static friction)

First, let's calculate W, W_parallel, and W_perpendicular:

W = m * g
W_parallel = W * sinθ
W_perpendicular = W * cosθ

Then, we can calculate N:

N = W_perpendicular

Using the value of N, we can find Ff:

Ff = μs * N

Finally, we can solve for Δx:

Δx = (μs * N) / k

Substituting the values, we can calculate the maximum amount the spring can be stretched and the box remain at rest.

M*g = 2kg * 9.8N./kg = 19.6 N. = Wt. of

box.

Fp = 19.6*sin65 = 17.76 N. = Force
parallel to the incline.

Fn = 19.6*Cos65 = 8.28 N.=Perpendicular
to the incline = Normal.

Fs = u*Fn = 0.3 * 8.28 = 2.48 N. = Force
of static friction.

Fap-Fp-Fs = M*a
Fap-17.76-2.48 = M*0 = 0
Fap = 20.24 N. = Force applied.

17 * m = 20.24 N.
m = 1.19 m. = Max. distance.