A 8.6 kg board is wedged into a corner and held by a spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 186 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length.

To find the amount by which the spring is stretched, we need to consider the forces acting on the board and apply Newton's laws of motion.

First, let's analyze the forces acting on the board. We have the force of gravity pulling down vertically, the normal force pushing up perpendicular to the surface, and the force exerted by the spring.

1. Calculate the weight of the board: Fg = m * g, where m is the mass (8.6 kg) and g is the acceleration due to gravity (9.8 m/s²).
Fg = 8.6 kg * 9.8 m/s² = 84.28 N

2. Resolve the weight into vertical and horizontal components. Since the board is at a 50.0° angle, the vertical component is given by: Fv = Fg * sin(50.0°)
Fv = 84.28 N * sin(50.0°) = 64.57 N

3. The spring force acts parallel to the floor and opposes the horizontal component of the weight. The horizontal component can be calculated using: Fh = Fg * cos(50.0°)
Fh = 84.28 N * cos(50.0°) = 53.91 N

4. The spring force is given by Hooke's law: Fs = -k * x, where k is the spring constant (186 N/m) and x is the displacement of the spring from its unstrained length.

5. Since the board is in equilibrium, the vertical and horizontal forces must balance. So, we have Fv = Fs.

6. Solve for the displacement of the spring: Fs = k * x
64.57 N = 186 N/m * x
x = 64.57 N / (186 N/m)
x ≈ 0.347 m

Therefore, the amount by which the spring is stretched from its unstrained length is approximately 0.347 meters.