To find the spring constant of the spring, we can use the principles of equilibrium for the block on the inclined plane.

Step 1: Analyzing Forces

The forces acting on the block are:

1. Gravitational force: $F_g = mg$, where $g$ is the acceleration due to gravity ($g \approx 9.8 \, \text{m/s}^2$).
2. Normal force: Perpendicular to the inclined plane.
3. Spring force: $F_s = kx$, acting in the direction opposite to the displacement of the spring.
4. Component of gravitational force parallel to the incline: $F_{\parallel} = mg \sin(\theta)$.

Step 2: Setting up the Equilibrium Equation

For the block to be in equilibrium on the inclined plane, the force due to the spring should balance the component of the gravitational force parallel to the incline: $kx = mg \sin(\theta)$

Step 3: Solve for the Spring Constant $k$

Rearrange the equation to solve for $k$: $k = \frac{mg \sin(\theta)}{x}$

Step 4: Substitute the Known Values

Given:

• $m = 0.55 \, \text{kg}$
• $g = 9.8 \, \text{m/s}^2$
• $\theta = 17^\circ$
• $x = 0.5 \, \text{m}$

Substituting these into the equation: $k = \frac{(0.55 \, \text{kg})(9.8 \, \text{m/s}^2) \sin(17^\circ)}{0.5 \, \text{m}}$

Calculate the value: $k = \frac{(0.55)(9.8)(0.2924)}{0.5} \approx \frac{1.576}{0.5} = 3.15 \, \text{N/m}$

Final Answer

The spring constant of the spring is: $k \approx 3.15 \, \text{N/m}$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How does the angle of inclination affect the force required to stretch the spring?
2. What happens to the spring constant if the mass of the block increases?
3. How would the equilibrium equation change if the inclined plane had friction?
4. What is the potential energy stored in the spring in this setup?
5. How would the spring constant change if the block was attached to two springs in parallel?

Tip: The spring constant $k$ measures how stiff the spring is. A higher value means a stiffer spring.