Let's solve this step-by-step.

Part (a): Elastic Potential Energy of the Block-Spring System

The elastic potential energy ($U$) stored in a compressed or stretched spring is given by: $U = \frac{1}{2} k x^2$ where:

• $k$ is the spring constant (815 N/m),
• $x$ is the compression or extension of the spring (0.0750 m).

Substituting the values: $U = \frac{1}{2} \times 815 \, \text{N/m} \times (0.0750 \, \text{m})^2$

Calculating this: $U = \frac{1}{2} \times 815 \times 0.005625 = 2.2903 \, \text{J}$

So, the elastic potential energy of the system is approximately: $U \approx 2.29 \, \text{J}$

Part (b): Speed of the Block After Leaving the Spring

Since the surface is frictionless, the elastic potential energy of the spring will convert entirely into the kinetic energy of the block once it leaves the spring.

The kinetic energy ($K$) of the block is given by: $K = \frac{1}{2} m v^2$ where:

• $m$ is the mass of the block (3.80 kg),
• $v$ is the speed of the block.

Since all the potential energy is converted to kinetic energy: $U = K$ Thus, $2.29 = \frac{1}{2} \times 3.80 \times v^2$

Rearrange to solve for $v$: $v^2 = \frac{2 \times 2.29}{3.80}$ $v^2 = \frac{4.58}{3.80} \approx 1.2053$ $v \approx \sqrt{1.2053} \approx 1.10 \, \text{m/s}$

So, the speed of the block after leaving the spring is approximately: $v \approx 1.10 \, \text{m/s}$

Summary:

• (a) Elastic potential energy: $2.29 \, \text{J}$
• (b) Speed of the block: $1.10 \, \text{m/s}$

Would you like more details on any part of the solution or have any questions?

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Here are five related questions for further understanding:

1. How would the speed change if the mass of the block were doubled?
2. What would happen if there was friction on the surface?
3. How does the spring constant affect the energy stored in the spring?
4. What is the relationship between kinetic and potential energy in this scenario?
5. How would the result differ if the block was initially compressed on an inclined plane?

Tip: For springs, the energy is proportional to the square of the compression or extension, so doubling the compression increases the energy fourfold.