To solve the given problem, we will use the formulas for elastic potential energy and the relationship between potential energy and kinetic energy.
(a) Elastic Potential Energy
The elastic potential energy \( U \) stored in a compressed or stretched spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where:
- \( k \) is the spring constant,
- \( x \) is the compression or extension of the spring.
Given:
- \( k = 855 , \text{N/m} \)
- \( x = 0.0400 , \text{m} \)
Substituting these values into the formula:
\[ U = \frac{1}{2} \times 855 , \text{N/m} \times (0.0400 , \text{m})^2 \]
Calculating this:
\[ U = \frac{1}{2} \times 855 \times 0.0016 , \text{m}^2 = \frac{1}{2} \times 1.368 , \text{J} \approx 0.684 , \text{J} \]
Thus, the elastic potential energy of the block-spring system is:
\[ \boxed{0.684 , \text{J}} \]
(b) Block's Speed After Leaving the Spring
When the block is released and the surface is frictionless, all the elastic potential energy will convert into kinetic energy. The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 \]
Setting the kinetic energy equal to the potential energy:
\[ \frac{1}{2} mv^2 = U \]
Substituting the values we know:
- \( m = 2.00 , \text{kg} \)
- \( U = 0.684 , \text{J} \)
Thus,
\[ \frac{1}{2} \times 2.00 , \text{kg} \times v^2 = 0.684 , \text{J} \]
This simplifies to:
\[ 1.00 , \text{kg} \times v^2 = 0.684 , \text{J} \]
Solving for \( v^2 \):
\[ v^2 = 0.684 , \text{m}^2/\text{s}^2 \]
Taking the square root to find \( v \):
\[ v = \sqrt{0.684} \approx 0.826 , \text{m/s} \]
Thus, the speed of the block after leaving the spring is:
\[ \boxed{0.826 , \text{m/s}} \]
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