A block of mass 2 kg moving on a horizontal surface with speed of `4ms^(-1)` enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by `F=-kx` where `k=12Nm^(-1)`. The speed of the block as it just crosses the rough surface will be :

Options

A. zero
B. `1.5ms^(-1)`
C. `2.0ms^(-1)`
D. `2.5ms^(-1)`

Explanation

To solve the problem, we will use the work-energy principle. The work done by the retarding force will equal the change in kinetic energy of the block.

### Step 1: Identify the initial conditions
The block has a mass \( m = 2 \, \text{kg} \) and an initial speed \( u = 4 \, \text{m/s} \). The block enters the rough surface at \( x = 0.5 \, \text{m} \) and exits at \( x = 1.5 \, \text{m} \).

### Step 2: Determine the retarding force
The retarding force is given by:
\[ F = -kx \]
where \( k = 12 \, \text{N/m} \). Therefore, the force varies with position \( x \).

### Step 3: Calculate the work done by the retarding force
The work done \( W \) by the retarding force as the block moves from \( x = 0.5 \, \text{m} \) to \( x = 1.5 \, \text{m} \) is given by:
\[ W = \int_{x_1}^{x_2} F \, dx = \int_{0.5}^{1.5} -kx \, dx \]
Substituting \( k = 12 \):
\[ W = \int_{0.5}^{1.5} -12x \, dx \]

### Step 4: Evaluate the integral
Calculating the integral:
\[ W = -12 \left[ \frac{x^2}{2} \right]_{0.5}^{1.5} = -12 \left( \frac{(1.5)^2}{2} - \frac{(0.5)^2}{2} \right) \]
Calculating the squares:
\[ = -12 \left( \frac{2.25}{2} - \frac{0.25}{2} \right) = -12 \left( \frac{2.25 - 0.25}{2} \right) = -12 \left( \frac{2}{2} \right) = -12 \]
Thus, the work done by the retarding force is \( W = -12 \, \text{J} \).

### Step 5: Calculate the change in kinetic energy
The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \, \text{J} \]
Let \( v \) be the final speed of the block after crossing the rough surface. The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} m v^2 \]
According to the work-energy principle:
\[ W = KE_f - KE_i \]
Substituting the values:
\[ -12 = \frac{1}{2} \times 2 v^2 - 16 \]
This simplifies to:
\[ -12 = v^2 - 16 \]
Rearranging gives:
\[ v^2 = 4 \]
Taking the square root:
\[ v = 2 \, \text{m/s} \]

### Final Answer
The speed of the block as it just crosses the rough surface is \( \boxed{2 \, \text{m/s}} \).

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