As shown in the figure, a 1.50-kg box is held at rest against a spring with a force constant *k* = 715 N/m that is compressed a distance *d*. When the box is released, it slides across a surface that is frictionless, except for a rough patch that has a coefficient of kinetic friction 𝜇*k* = 0.400 and is 6.00 cm in length. If the speed of the box is 1.90 m/s after sliding across the rough patch, determine the initial compression *d* of the spring.
 cm

To determine the initial compression \( d \) of the spring, we can use the principle of energy conservation, considering the potential energy stored in the spring and the work done against friction as the box slides across the rough patch.

1. **Potential Energy in the Spring:**
   The potential energy \( PE \) stored in the spring when compressed is given by:
   \[
   PE = \frac{1}{2} k d^2
   \]
   where \( k = 715 \, \text{N/m} \) and \( d \) is the compression.

2. **Kinetic Energy after the Rough Patch:**
   The kinetic energy \( KE \) of the box after it has slid across the rough patch is:
   \[
   KE = \frac{1}{2} m v^2
   \]
   where \( m = 1.50 \, \text{kg} \) and \( v = 1.90 \, \text{m/s} \).

3. **Work Done Against Friction:**
   The work done against friction \( W_f \) while sliding through the rough patch is given by:
   \[
   W_f = f_k \cdot d_{rough}
   \]
   where the kinetic friction force \( f_k \) is:
   \[
   f_k = \mu_k m g
   \]
   Here, \( \mu_k = 0.400 \), \( g = 9.81 \, \text{m/s}^2 \), and \( d_{rough} = 0.06 \, \text{m} \).

### Step 1: Calculate the work done against friction

Calculating \( f_k \):
\[
f_k = 0.400 \times 1.50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 5.886 \, \text{N}
\]

Now, calculate \( W_f \):
\[
W_f = f_k \cdot d_{rough} = 5.886 \, \text{N} \times 0.06 \, \text{m} = 0.35316 \, \text{J}
\]

### Step 2: Calculate the kinetic energy after the rough patch

Calculating \( KE \):
\[
KE = \frac{1}{2} \times 1.50 \, \text{kg} \times (1.90 \, \text{m/s})^2 = \frac{1}{2} \times 1.50 \times 3.61 = 2.7115 \, \text{J}
\]

### Step 3: Apply the energy conservation principle

The initial potential energy in the spring is equal to the kinetic energy after sliding and the work done against friction:
\[
\frac{1}{2} k d^2 = KE + W_f
\]

Substituting the known values:
\[
\frac{1}{2} \times 715 \, \text{N/m} \times d^2 = 2.7115 \, \text{J} + 0.35316 \, \text{J}
\]

Simplifying:
\[
\frac{1}{2} \times 715 \, d^2 = 3.06466 \, \text{J}
\]
\[
715 \, d^2 = 6.12932
\]
\[
d^2 = \frac{6.12932}{715}
\]
\[
d^2 \approx 0.00856
\]
\[
d \approx \sqrt{0.00856} \approx 0.0925 \, \text{m} = 9.25 \, \text{cm}
\]

### Final Answer
The initial compression \( d \) of the spring is approximately **9.25 cm**.

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. How does the coefficient of friction affect the speed after the rough patch?
2. What would happen if the surface were completely frictionless?
3. How does changing the mass of the box influence the spring compression?
4. What is the potential energy change if the spring constant is doubled?
5. How would you calculate the work done if the rough patch were longer? 

Tip: Always check the units in your calculations to ensure consistency and accuracy!

As shown in the figure, a 1.50-kg box is held at rest against a spring with a force constant k = 715 N/m that is compressed a distance d. When the box is released, it slides across a surface that is frictionless, except for a rough patch that has a coefficient of kinetic friction μk = 0.400 and is 6.00 cm in length. If the speed of the box is 1.90 m/s after sliding across the rough patch, determine the initial compression d of the spring.

Explore how to calculate the initial compression of a spring in a physics problem involving energy conservation and friction. A 1.50-kg box slides over a rough patch, revealing important concepts in kinetic energy and potential energy. Learn the process and calculations involved.

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