Two block of masses `m_(1)` and `m_(2)` connected by a light spring rest on a horizontal plane. The coefficient of friction between the block and the surface is equal to `mu`. What minimum constant force has to be applied in the horizontal direction to the block of mass `m_(1)` in order to shift the other block?

Options

A. 8 N
B. 15 N
C. 10 N
D. 25 N

Answer

To solve the problem of determining the minimum constant force \( F \) that must be applied to block \( m_1 \) in order to shift block \( m_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
- Block \( m_1 \) is connected to block \( m_2 \) by a light spring. The coefficient of friction between both blocks and the surface is \( \mu \).
- The force \( F \) is applied to block \( m_1 \) in the horizontal direction.

2. **Identifying the Condition for Block \( m_2 \) to Move**:
- For block \( m_2 \) to start moving, the force exerted by the spring on block \( m_2 \) (let's denote it as \( kx \), where \( k \) is the spring constant and \( x \) is the extension of the spring) must overcome the frictional force acting on block \( m_2 \).
- The limiting frictional force \( f_{\text{friction}} \) on block \( m_2 \) is given by:
\[ f_{\text{friction}} = \mu m_2 g \]

3. **Setting Up the Equation**:
- For block \( m_2 \) to start moving, we need:
\[ kx \geq \mu m_2 g \]

4. **Applying Work-Energy Principle**:
- The work done by the applied force \( F \) must equal the work done against friction and the work done on the spring. The work-energy theorem states:
\[ F \cdot x = f_{\text{friction}} \cdot x + \frac{1}{2} k x^2 \]
- Substituting \( f_{\text{friction}} \):
\[ F \cdot x = \mu m_2 g \cdot x + \frac{1}{2} k x^2 \]

5. **Simplifying the Equation**:
- Dividing through by \( x \) (assuming \( x \neq 0 \)):
\[ F = \mu m_2 g + \frac{1}{2} k x \]

6. **Finding Minimum Force \( F \)**:
- To find the minimum force \( F \), we need to minimize the term \( kx \). The minimum value of \( kx \) occurs when \( kx = \mu m_2 g \) (from step 3).
- Substituting this back into the equation for \( F \):
\[ F_{\text{min}} = \mu m_2 g + \frac{1}{2} \mu m_2 g = \frac{3}{2} \mu m_2 g \]

7. **Final Expression**:
- Therefore, the minimum constant force \( F \) that must be applied to block \( m_1 \) to shift block \( m_2 \) is:
\[ F_{\text{min}} = \mu (m_1 + m_2) g \]