A m = 1.85 kg box rests on a plank that is inclined at an angle of θ = 65.5° above the horizontal. The upper end of the box is attached to a spring with a force constant of 21.0 N/m. If the maximum force of static friction between the box and the plank is 1.60 N, what is the maximum amount the spring can be stretched and the box remain at rest?

weight down plank: mg*SinTheta

Spring pull=kx^2
friction down: 1.6 (spring pulls up, friction opposes that)

spring pull-frictiondown=weight down plank
solve for x

To find the maximum amount the spring can be stretched while the box remains at rest, we need to determine the net force acting on the box parallel to the incline. Let's break it down step by step:

1. Draw a diagram: Sketch the inclined plane, with the box on top. Label the given values: mass of the box (m = 1.85 kg), angle of inclination (θ = 65.5°), force constant of the spring (k = 21.0 N/m), and maximum force of static friction (fs = 1.60 N).

2. Resolve forces: Break down the forces acting on the box parallel and perpendicular to the incline. The weight of the box (mg) can be resolved into two components: mg*sin(θ) acting parallel to the incline and mg*cos(θ) acting perpendicular to the incline. The normal force (N) acting perpendicular to the incline cancels out the mg*cos(θ) component.

3. Determine the maximum static friction force: The maximum static friction force (fs) is given as 1.60 N. Since the box is at rest, the static friction force balances the component of the weight parallel to the incline. Therefore, fs = mg*sin(θ).

4. Calculate the maximum displacement: The force exerted by the spring (fspring) opposes the maximum static friction force. Therefore, fspring = fs = mg*sin(θ). We can use Hooke's Law to relate the force exerted by the spring to the displacement (x) of the spring from its equilibrium position: fspring = k*x. Rearranging the equation, we find x = fspring / k.

Let's plug in the values and calculate the maximum amount the spring can be stretched:

x = fspring / k = (mg*sin(θ)) / k
= (1.85 kg * 9.8 m/s^2 * sin(65.5°)) / (21.0 N/m)
= 0.0886 m

Therefore, the maximum amount the spring can be stretched while the box remains at rest is 0.0886 meters.