A `0.5kg` block slides from the point A on a horizontal track with an initial speed `3 m//s` towards a weightless horizontal spring of length `1m` and force constant `2N//m.` The part AB of the track is frictionless and the part BC has the confficient of static and kinetic friction as `0.20` respectively. If the distancences AB and BD are `2m` and `2.14m` respectively, find total distance through which the block moves before it comes to rest completely. `(g=10 m//s^(2) ).

Explanation

To solve the problem step by step, we will analyze the motion of the block from point A to point D, taking into account the effects of friction and the spring.

### Step 1: Calculate the initial kinetic energy at point A
The initial kinetic energy (KE) of the block can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 0.5 \, \text{kg} \) (mass of the block)
- \( v = 3 \, \text{m/s} \) (initial speed)

Substituting the values:
\[ KE = \frac{1}{2} \times 0.5 \times (3)^2 = \frac{1}{2} \times 0.5 \times 9 = 2.25 \, \text{J} \]

### Step 2: Calculate the work done against friction on path BD
The work done against friction can be calculated using the formula:
\[ W = \mu_k m g d \]
Where:
- \( \mu_k = 0.2 \) (coefficient of kinetic friction)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( d = 2.14 \, \text{m} \) (distance on path BD)

Substituting the values:
\[ W = 0.2 \times 0.5 \times 10 \times 2.14 = 2.14 \, \text{J} \]

### Step 3: Calculate the kinetic energy at point D
The kinetic energy at point D can be found by subtracting the work done against friction from the initial kinetic energy:
\[ KE_D = KE - W = 2.25 \, \text{J} - 2.14 \, \text{J} = 0.11 \, \text{J} \]

### Step 4: Determine the compression of the spring
When the block compresses the spring, the kinetic energy will be converted into spring potential energy and work done against friction. The energy conservation equation can be written as:
\[ KE_D = \frac{1}{2} k x^2 + \mu_k m g x \]
Where:
- \( k = 2 \, \text{N/m} \) (spring constant)
- \( x \) is the compression of the spring.

Substituting the known values:
\[ 0.11 = \frac{1}{2} \times 2 \times x^2 + 0.2 \times 0.5 \times 10 \times x \]
This simplifies to:
\[ 0.11 = x^2 + x \]
Rearranging gives:
\[ x^2 + x - 0.11 = 0 \]

### Step 5: Solve the quadratic equation for x
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1, c = -0.11 \).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-0.11)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 0.44}}{2} = \frac{-1 \pm \sqrt{1.44}}{2} \]
Calculating the square root:
\[ \sqrt{1.44} = 1.2 \]
Thus:
\[ x = \frac{-1 + 1.2}{2} = \frac{0.2}{2} = 0.1 \, \text{m} \]
(The negative root is not physically meaningful in this context.)

### Step 6: Calculate the total distance moved by the block
The total distance moved by the block is the sum of the distances AB, BD, and the compression of the spring:
\[ \text{Total distance} = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 \, \text{m} \]

### Final Answer
The total distance through which the block moves before it comes to rest completely is **4.24 m**.

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