To find the speed of the book as it slides away, we can use the principle of energy conservation. When the student compresses the spring, it stores potential energy, which is then transferred to the book as kinetic energy when the spring is released.

Step-by-Step Solution

1. Convert the Mass and Compression to SI Units:

   • Mass of the book: $m = 490 \ \text{g} = 0.490 \ \text{kg}$
   • Compression of the spring: $x = 9.80 \ \text{cm} = 0.098 \ \text{m}$

2. Calculate the Potential Energy Stored in the Spring: The potential energy $U$ stored in a compressed (or stretched) spring is given by: $U = \frac{1}{2} k x^2$ where $k = 1550 \ \text{N/m}$ is the spring constant, and $x = 0.098 \ \text{m}$ is the compression.

   Plugging in the values: $U = \frac{1}{2} \times 1550 \times (0.098)^2$ $U = \frac{1}{2} \times 1550 \times 0.009604$ $U = 7.4426 \ \text{J}$

3. Apply Energy Conservation: The potential energy in the spring is converted to the kinetic energy of the book as it slides away: $U = \frac{1}{2} m v^2$ where $v$ is the speed of the book.

4. Solve for the Speed $v$: $v = \sqrt{\frac{2U}{m}}$ Substituting $U = 7.4426 \ \text{J}$ and $m = 0.490 \ \text{kg}$: $v = \sqrt{\frac{2 \times 7.4426}{0.490}}$ $v = \sqrt{\frac{14.8852}{0.490}}$ $v = \sqrt{30.3782}$ $v \approx 5.51 \ \text{m/s}$

Final Answer:

The speed of the book as it slides away is approximately $\mathbf{5.51 \ m/s}$.

Would you like further details on any part of this solution?

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1. How is energy conserved in spring and kinetic energy scenarios?
2. How does spring compression affect potential energy?
3. What changes if friction is added to this system?
4. How does changing the spring constant affect speed?
5. What role does mass play in determining speed?

Tip: In spring problems, always verify units, especially for displacement and mass, to ensure correct calculations.