To solve for the block's speed when the spring returns to its uncompressed length, we can use the concept of energy conservation.
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Calculate the potential energy stored in the compressed spring: The potential energy \( U \) stored in a compressed or stretched spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where:
- \( k = 200 , \text{N/m} \) (the spring constant),
- \( x = 0.1 , \text{m} \) (the compression of the spring).
Plugging in the values:
\[ U = \frac{1}{2} (200 , \text{N/m}) (0.1 , \text{m})^2 = \frac{1}{2} (200) (0.01) = 1 , \text{J} \]
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Use energy conservation to find the kinetic energy of the block: When the spring pushes the block back to its uncompressed position, all the potential energy stored in the spring is converted into kinetic energy \( K \) of the block (since there are no other forces doing work, such as friction).
The kinetic energy \( K \) of the block is given by the formula:
\[ K = \frac{1}{2} mv^2 \]
where \( m = 4 , \text{kg} \) is the mass of the block, and \( v \) is its speed when the spring is uncompressed. According to the conservation of energy:
\[ U = K \]
Thus,
\[ 1 , \text{J} = \frac{1}{2} (4 , \text{kg}) v^2 \]
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Solve for the speed \( v \): Rearranging the equation to solve for \( v^2 \):
\[ 1 = 2 v^2 \quad \Rightarrow \quad v^2 = \frac{1}{2} \]
Taking the square root of both sides:
\[ v = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 , \text{m/s} \]
Thus, the speed of the block when the spring returns to its uncompressed length is approximately 0.707 m/s.