A block of mass 1 kg is attached to one end of a spring of force constant k = 20 N/m. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface `(mu = 0.04)`. The initial displacement of the block from the equilibrium position is a = 30 cm. How many times the block passes from the mean position before coming to rest ? `(g = 10 m//s^(2))`

Options

A. 11
B. 7
C. 6
D. 15

Answer

To solve the problem of how many times the block passes through the mean position before coming to rest, we will follow these steps:

### Step 1: Understand the System
We have a block of mass \( m = 1 \, \text{kg} \) attached to a spring with spring constant \( k = 20 \, \text{N/m} \). The block is on a rough horizontal surface with a coefficient of friction \( \mu = 0.04 \). The initial displacement (amplitude) of the block from the equilibrium position is \( A = 30 \, \text{cm} = 0.3 \, \text{m} \).

### Step 2: Calculate the Force of Friction
The force of friction \( F_f \) acting on the block can be calculated using the formula:
\[ F_f = \mu \cdot m \cdot g \]
Substituting the values:
\[ F_f = 0.04 \cdot 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.4 \, \text{N} \]

### Step 3: Calculate the Potential Energy in the Spring
The potential energy \( PE \) stored in the spring when the block is at maximum displacement (amplitude) is given by:
\[ PE = \frac{1}{2} k A^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \cdot 20 \, \text{N/m} \cdot (0.3 \, \text{m})^2 = \frac{1}{2} \cdot 20 \cdot 0.09 = 0.09 \, \text{J} \]

### Step 4: Work Done by Friction
The work done by friction \( W_f \) when the block moves from the maximum displacement to the mean position is:
\[ W_f = F_f \cdot d \]
where \( d \) is the distance moved. The block moves a distance of \( A \) to reach the mean position, so:
\[ W_f = 0.4 \, \text{N} \cdot 0.3 \, \text{m} = 0.12 \, \text{J} \]

### Step 5: Determine the Energy Loss per Oscillation
As the block oscillates, it loses energy due to friction. The energy lost in each complete oscillation (to the mean position and back) can be calculated as:
\[ \text{Energy lost} = 2 \cdot W_f = 2 \cdot 0.12 \, \text{J} = 0.24 \, \text{J} \]

### Step 6: Calculate the Number of Oscillations
The total potential energy at the start is \( 0.09 \, \text{J} \). The number of oscillations \( n \) before the energy is used up can be calculated by:
\[ n = \frac{\text{Initial Potential Energy}}{\text{Energy lost per oscillation}} = \frac{0.09 \, \text{J}}{0.24 \, \text{J}} \approx 0.375 \]
Since the block cannot complete a fraction of an oscillation, we take the integer part, which is 0.

### Step 7: Conclusion
The block does not complete any full oscillations before coming to rest. Therefore, it passes through the mean position **0 times** before coming to rest.