This problem involves a block and a spring, and we need to calculate several things based on energy principles and spring mechanics. Here's a step-by-step solution:

Given:

• Mass of block $m = 872 \, \text{g} = 0.872 \, \text{kg}$
• Spring constant $k = 460 \, \text{N/m}$
• Compression of spring $x = 23.1 \, \text{cm} = 0.231 \, \text{m}$

We'll tackle the problem part by part:

(a) How much work is done by the block on the spring?

The work done by the block on the spring is equivalent to the potential energy stored in the compressed spring. This is given by the formula:

$W_{\text{block}} = \frac{1}{2} k x^2$

Substituting the values:

$W_{\text{block}} = \frac{1}{2} (460 \, \text{N/m}) (0.231 \, \text{m})^2$ $W_{\text{block}} = \frac{1}{2} \times 460 \times 0.053361$ $W_{\text{block}} = 12.28 \, \text{J}$

(b) How much work is done by the spring on the block?

By Newton's third law, the work done by the spring on the block is the same in magnitude but opposite in direction as the work done by the block on the spring. Therefore:

$W_{\text{spring}} = -W_{\text{block}} = -12.28 \, \text{J}$

(c) What is the value of $h_0$?

To find the initial height $h_0$, we use the conservation of mechanical energy. Initially, the block has only gravitational potential energy, and when it compresses the spring, all this energy is converted into spring potential energy.

Using energy conservation:

$mgh_0 = \frac{1}{2} k x^2$

Solving for $h_0$:

$h_0 = \frac{\frac{1}{2} k x^2}{mg}$

Substituting the values:

$h_0 = \frac{\frac{1}{2} \times 460 \times (0.231)^2}{0.872 \times 9.81}$ $h_0 = \frac{12.28}{8.54832}$ $h_0 \approx 1.44 \, \text{m}$

(d) What would be the maximum compression of the spring if the block were released from height $3h_0$?

If the block is released from height $3h_0$, it would have three times the gravitational potential energy. This energy would be converted into spring potential energy at maximum compression.

Using energy conservation again:

$mg(3h_0) = \frac{1}{2} k x_{\text{max}}^2$

We know $h_0$ from part (c), so:

$mgh_0 = \frac{1}{2} k x^2$

Thus:

$3mgh_0 = \frac{1}{2} k x_{\text{max}}^2$

Substituting $h_0 = 1.44 \, \text{m}$:

$3 \times 0.872 \times 9.81 \times 1.44 = \frac{1}{2} \times 460 \times x_{\text{max}}^2$

Simplifying:

$36.88 = 230 x_{\text{max}}^2$

Solving for $x_{\text{max}}$:

$x_{\text{max}}^2 = \frac{36.88}{230} = 0.1603$

$x_{\text{max}} = \sqrt{0.1603} \approx 0.4 \, \text{m}$

Thus, the maximum compression of the spring is approximately 0.4 meters.

Would you like further clarification on any of these steps or details? Here are some related questions for further exploration:

1. How would the work done change if the block's mass were doubled?
2. What would happen to the spring compression if the spring constant were halved?
3. How does the potential energy of the spring vary with the compression distance?
4. What is the importance of energy conservation in spring-block systems?
5. How would air resistance affect the energy conversions in this scenario?

Tip: Always ensure the units in your calculations are consistent, especially when dealing with different forms of energy (e.g., gravitational vs. spring potential energy).