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We are given the following information in this problem:

- Mass of the block \( m = 2.5 \, \text{kg} \)
- Spring constant \( k = 70 \, \text{N/m} \)
- Angle of the incline \( \theta = 35^\circ \)
- The incline is smooth and frictionless.

We need to find the distance \( x \) that the block moves down the incline before momentarily coming to rest (this happens when the potential energy of the block equals the potential energy stored in the spring).

### Step-by-step solution:

1. **Work-Energy Principle**:
   The total mechanical energy is conserved. Initially, the block only has gravitational potential energy, which gets converted into spring potential energy when the block moves down the incline.

2. **Gravitational Potential Energy (U)**:
   The gravitational potential energy lost when the block moves down a distance \( x \) along the incline is:
   \[
   \Delta U = m g x \sin\theta
   \]
   where \( g = 9.8 \, \text{m/s}^2 \).

3. **Spring Potential Energy**:
   The potential energy stored in the spring when compressed by a distance \( x \) is:
   \[
   U_{\text{spring}} = \frac{1}{2} k x^2
   \]

4. **Energy Conservation**:
   The gravitational potential energy lost by the block is equal to the potential energy stored in the spring:
   \[
   m g x \sin\theta = \frac{1}{2} k x^2
   \]

5. **Substitute values**:
   Substituting the given values into the energy equation:
   \[
   2.5 \times 9.8 \times x \times \sin(35^\circ) = \frac{1}{2} \times 70 \times x^2
   \]

6. **Simplify the equation**:
   First, calculate \( \sin(35^\circ) \approx 0.574 \):
   \[
   2.5 \times 9.8 \times 0.574 \times x = 35 x^2
   \]
   Simplifying further:
   \[
   14.105 x = 35 x^2
   \]

7. **Solve for \( x \)**:
   Divide both sides by \( x \) (assuming \( x \neq 0 \)):
   \[
   14.105 = 35 x
   \]
   Solving for \( x \):
   \[
   x = \frac{14.105}{35} \approx 0.403 \, \text{m}
   \]

So, the block moves approximately \( 0.40 \, \text{m} \) down the incline.

The correct answer is:
- **c) 0.40 m**

Would you like more details, or do you have any questions about the process?

### Related Questions:
1. What would happen to the distance if the spring constant were doubled?
2. How would the answer change if the angle of the incline were steeper?
3. What if friction were introduced—how would we modify the energy conservation equation?
4. What other forces could act on the block if the incline were not frictionless?
5. How do different types of energy (kinetic, potential) influence the motion of a block attached to a spring?

**Tip**: When solving energy conservation problems, always consider all forms of energy present in the system to ensure total mechanical energy is conserved.

A block has a mass m = 2.5 kg situated on a smooth frictionless incline is connected to a spring of negligible mass, the spring constant (k = 70 N/m). The block is released from rest when the spring is unstretched. The pulley is frictionless and massless. The angle (θ) between the incline and horizontal is 35°. The distance (in m) the block will move down the incline coming (momentarily) to rest is: a. 0.23, b. 0.34, c. 0.40, d. 0.46, e. 0.49.

Solve an energy conservation problem involving a 2.5 kg block on a smooth incline connected to a spring with a constant of 70 N/m. The block is released from rest and moves a distance down the incline before coming to rest. Explore key concepts like gravitational and spring potential energy.

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