An ice cube is kept on an inclined plane of angle `30^(@)`. The coefficient to kinetic friction between the block and incline plane is `1//sqrt(3)`. What is the acceleration of the block ?

Options

A. Zero
B. `2 m//s^(2)`
C. `1.5 m//s^(2)`
D. `5 m//s^(2)`

Explanation

To find the acceleration of the ice cube on the inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the ice cube
The forces acting on the ice cube are:
1. The gravitational force (\(mg\)) acting downwards.
2. The normal force (\(N\)) acting perpendicular to the inclined plane.
3. The frictional force (\(F_f\)) acting opposite to the direction of motion along the plane.

### Step 2: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
- Parallel to the incline: \(F_{\text{parallel}} = mg \sin \theta\)
- Perpendicular to the incline: \(F_{\text{perpendicular}} = mg \cos \theta\)

### Step 3: Calculate the frictional force
The frictional force can be calculated using the coefficient of kinetic friction (\(\mu\)) and the normal force (\(N\)):
\[ F_f = \mu N \]
Since \(N = mg \cos \theta\), we have:
\[ F_f = \mu (mg \cos \theta) \]

### Step 4: Write the equation of motion
The net force acting on the ice cube along the incline is given by:
\[ F_{\text{net}} = F_{\text{parallel}} - F_f \]
Substituting the expressions we derived:
\[ F_{\text{net}} = mg \sin \theta - \mu (mg \cos \theta) \]

### Step 5: Apply Newton's second law
According to Newton's second law, the net force is equal to mass times acceleration (\(ma\)):
\[ mg \sin \theta - \mu (mg \cos \theta) = ma \]
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ g \sin \theta - \mu g \cos \theta = a \]

### Step 6: Substitute the known values
Given:
- \(\theta = 30^\circ\)
- \(\mu = \frac{1}{\sqrt{3}}\)
- \(g = 10 \, \text{m/s}^2\)

Now substitute these values into the equation:
\[ a = g \sin(30^\circ) - \mu g \cos(30^\circ) \]

Calculating the trigonometric values:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)

Substituting these values:
\[ a = 10 \left(\frac{1}{2}\right) - \frac{1}{\sqrt{3}} \cdot 10 \left(\frac{\sqrt{3}}{2}\right) \]
\[ a = 5 - 5 = 0 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the ice cube on the inclined plane is \(0 \, \text{m/s}^2\).

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