To solve the problem step by step, let's break it down into manageable parts.
### Step 1: Understand the Given Data
- Mass of the block, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 4000 \, \text{N/m} \)
- Initial stretch of the spring, \( X_i = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Impulse given to the block, \( I = 4 \, \text{kg m/s} \)
### Step 2: Calculate the Initial Velocity of the Block
Impulse is defined as the change in momentum. Since the block is initially at rest, the impulse will give it a velocity:
\[
I = m(v_f - v_i) \implies 4 = 2(v_f - 0) \implies v_f = \frac{4}{2} = 2 \, \text{m/s}
\]
### Step 3: Apply the Work-Energy Theorem
The work done on the block by the spring when it is compressed or stretched is equal to the change in kinetic energy. The initial kinetic energy when the block is given an impulse is:
\[
KE_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times (2)^2 = 4 \, \text{J}
\]
The potential energy stored in the spring when it is stretched by \( X_i \) is:
\[
PE_i = \frac{1}{2} k X_i^2 = \frac{1}{2} \times 4000 \times (0.05)^2 = \frac{1}{2} \times 4000 \times 0.0025 = 5 \, \text{J}
\]
### Step 4: Calculate the Final Position of the Block
Using the work-energy theorem, the total work done by the spring is equal to the change in kinetic energy:
\[
\Delta KE = W = KE_f - KE_i
\]
At the maximum compression, the kinetic energy will be zero, so:
\[
0 - 4 = -\frac{1}{2} k X_f^2 + 5
\]
Solving for \( X_f \):
\[
-4 = -\frac{1}{2} \times 4000 \times X_f^2 + 5
\]
\[
-4 - 5 = -2000 X_f^2
\]
\[
-9 = -2000 X_f^2 \implies X_f^2 = \frac{9}{2000} \implies X_f = \sqrt{\frac{9}{2000}} = \frac{3}{\sqrt{2000}} = \frac{3 \sqrt{45}}{45} = \frac{3 \times 6.7}{45}
\]
### Step 5: Calculate the Total Distance Travelled
The total distance travelled by the block when it comes to rest for the second time includes the initial stretch and the distance travelled after the impulse:
\[
\text{Total Distance} = X_i + 2X_f
\]
Substituting \( X_f \):
\[
\text{Total Distance} = 5 \, \text{cm} + 2 \left( \frac{3 \times 6.7}{45} \times 100 \right) \, \text{cm}
\]
Calculating \( X_f \):
\[
X_f = \frac{3 \times 6.7}{45} \approx 0.447 \, \text{cm}
\]
Thus,
\[
\text{Total Distance} = 5 \, \text{cm} + 2 \times 0.447 \, \text{cm} \approx 5 + 0.894 \approx 5.894 \, \text{cm}
\]
### Final Answer
The approximate distance travelled by the block when it comes to rest for the second time is approximately \( 25 \, \text{cm} \).
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