To solve for the block's speed after it leaves the spring, we can use the concept of energy conservation. Here's how:

1. Elastic Potential Energy in the Spring: When the spring is compressed, it stores potential energy. This energy will convert entirely into the block's kinetic energy when the spring releases it (assuming a frictionless surface).

2. Formula for Elastic Potential Energy: The potential energy stored in the spring can be calculated using: $PE_{\text{spring}} = \frac{1}{2} k x^2$ where:

   • $k = 815 \, \text{N/m}$ is the spring constant,
   • $x = 0.0750 \, \text{m}$ is the compression distance.

3. Calculate Potential Energy: $PE_{\text{spring}} = \frac{1}{2} \times 815 \times (0.0750)^2$ $PE_{\text{spring}} = \frac{1}{2} \times 815 \times 0.005625$ $PE_{\text{spring}} = 2.29 \, \text{J}$

4. Conversion to Kinetic Energy: When the block leaves the spring, this potential energy becomes the block's kinetic energy: $KE = \frac{1}{2} m v^2$ where:

   • $m = 3.80 \, \text{kg}$ is the mass of the block,
   • $v$ is the speed of the block after leaving the spring.

5. Set Potential Energy Equal to Kinetic Energy: $2.29 = \frac{1}{2} \times 3.80 \times v^2$

6. Solve for $v$: $v^2 = \frac{2 \times 2.29}{3.80}$ $v^2 = \frac{4.58}{3.80}$ $v^2 = 1.205$ $v = \sqrt{1.205}$ $v \approx 1.10 \, \text{m/s}$

Final Answer:

The block's speed after leaving the spring is approximately $1.10 \, \text{m/s}$.

Would you like a more detailed explanation of any step?

Here are some related questions that might be helpful:

1. What would happen to the block’s speed if the spring constant increased?
2. How would friction affect the block's final speed?
3. What if the mass of the block were doubled?
4. How would the final speed change if the spring compression distance was doubled?
5. How can energy conservation help in problems involving gravitational potential energy?

Tip: In physics problems, always check if energy conservation or Newton's laws are more useful based on whether forces or energy transfer are emphasized.