Solution

Part (a) - Speed of the block at the moment the spring is no longer compressed

1. Conservation of Energy:
   The potential energy stored in the compressed spring is converted into the kinetic energy of the block.

   • Spring Potential Energy:
     $U_s = \frac{1}{2}kx^2$ where:

     • $k = 800 \, \mathrm{N/m}$ (spring constant),
     • $x = 0.03 \, \mathrm{m}$ (compression distance).

   • Kinetic Energy:
     $K = \frac{1}{2}mv^2$ where:

     • $m = 2.5 \, \mathrm{kg}$ (mass of the block),
     • $v$ is the velocity of the block.

   Using energy conservation: $U_s = K \implies \frac{1}{2}kx^2 = \frac{1}{2}mv^2$

2. Substitute values: $\frac{1}{2}(800)(0.03)^2 = \frac{1}{2}(2.5)v^2$

3. Simplify: $(400)(0.0009) = (1.25)v^2$ $0.36 = 1.25v^2$

4. Solve for $v$: $v^2 = \frac{0.36}{1.25} = 0.288$ $v = \sqrt{0.288} \approx 0.537 \, \mathrm{m/s}$

Thus, the speed of the block is approximately: $\boxed{0.537 \, \mathrm{m/s}}$

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Part (b) - Speed of the sphere when the spring is no longer compressed

1. Energy Conversion:
   For the sphere, the spring's potential energy is converted into both translational kinetic energy and rotational kinetic energy.

   • Translational Kinetic Energy:
     $K_t = \frac{1}{2}mv^2$

   • Rotational Kinetic Energy:
     $K_r = \frac{1}{2}I\omega^2$ where $I = \frac{2}{5}mR^2$ (moment of inertia of the sphere) and $\omega = \frac{v}{R}$ (angular velocity).

2. Total Energy Conservation: $U_s = K_t + K_r$ Substituting expressions: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2$

3. Simplify: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$ $\frac{1}{2}kx^2 = \left(\frac{1}{2} + \frac{1}{5}\right)mv^2$ $\frac{1}{2}kx^2 = \frac{7}{10}mv^2$

4. Solve for $v$: $v^2 = \frac{\frac{1}{2}kx^2}{\frac{7}{10}m}$ $v^2 = \frac{(0.5)(800)(0.03)^2}{0.7 \cdot 2.5}$ $v^2 = \frac{(400)(0.0009)}{1.75}$ $v^2 = \frac{0.36}{1.75} \approx 0.2057$ $v = \sqrt{0.2057} \approx 0.454 \, \mathrm{m/s}$

Thus, the linear speed of the sphere is approximately: $\boxed{0.454 \, \mathrm{m/s}}$

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Summary:

• (a) Speed of the block: $\boxed{0.537 \, \mathrm{m/s}}$
• (b) Speed of the sphere: $\boxed{0.454 \, \mathrm{m/s}}$

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Do you want further details or have any questions?
Here are 5 related questions to expand on this topic:

1. How would friction affect the results in this scenario?
2. What if the spring compression distance were doubled—how would it change the velocities?
3. How is angular momentum conserved in the second case with the sphere?
4. How would the result differ if the object were a hollow sphere instead of a solid one?
5. What is the energy lost due to rotational motion in the sphere case?

Tip: Always check if the system involves rotational motion when solving conservation of energy problems with objects that can roll or spin!