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 analyze the forces acting on the board and apply Hooke's law.
First, let's determine the gravitational force acting on the board. The weight of the board is given by the equation F_gravity = m * g, where m is the mass of the board and g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, m = 8.6 kg, so the weight of the board is F_gravity = 8.6 kg * 9.8 m/s^2 = 84.28 N.
Next, let's determine the horizontal component of the spring force acting on the board. The spring force can be calculated using the equation F_spring = k * x, where k is the spring constant and x is the displacement or stretch of the spring. In this case, we need to consider the horizontal component of the spring force, which is F_spring_horizontal = F_spring * cos(theta), where theta is the angle at which the spring is held (50.0°) and cos(theta) is the cosine of that angle. The cosine of 50.0° is approximately 0.643, so F_spring_horizontal = 186 N/m * x * 0.643.
Finally, we set up an equation by considering the equilibrium of forces in the vertical direction. The vertical component of the spring force must balance the weight of the board. Since the board is wedged in a corner and held at an angle, the vertical component of the spring force can be calculated using the equation F_spring_vertical = F_spring * sin(theta), where sin(theta) is the sine of the angle at which the spring is held. In this case, the sine of 50.0° is approximately 0.766, so F_spring_vertical = 186 N/m * x * 0.766. This must be equal to the weight of the board, so we have the equation:
186 N/m * x * 0.766 = 84.28 N
Now we can solve for x, which is the amount by which the spring is stretched. Rearranging the equation:
x = 84.28 N / (186 N/m * 0.766)
x ≈ 0.763 meters
Therefore, the spring is stretched by approximately 0.763 meters from its unstrained length.