A block placed on an inclined plane making an angle `17^(@)` with the horizontal slides down without any acceleration. What is the acceleration of the block if the inclimation is increased to `30^(@)` ?

Explanation

To solve the problem step by step, we need to analyze the forces acting on the block on the inclined plane and calculate the acceleration when the angle of inclination is increased from \(17^\circ\) to \(30^\circ\).

### Step 1: Understand the Forces Acting on the Block
When the block is on the inclined plane at an angle of \(17^\circ\), it is sliding down without any acceleration. This means that the forces acting on the block are balanced. The forces acting on the block include:
- The gravitational force \(mg\) acting downwards.
- The normal force \(R\) acting perpendicular to the inclined plane.
- The frictional force \(F_r\) acting up the incline.

### Step 2: Calculate the Forces at \(17^\circ\)
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \(mg \cos(17^\circ)\)
- Parallel to the incline: \(mg \sin(17^\circ)\)

Since the block is in equilibrium (no acceleration), we can write:
\[ mg \sin(17^\circ) = F_r \]
Where the frictional force \(F_r\) is given by:
\[ F_r = \mu R = \mu mg \cos(17^\circ) \]

Setting these equal gives:
\[ mg \sin(17^\circ) = \mu mg \cos(17^\circ) \]

Dividing both sides by \(mg\) (assuming \(m \neq 0\)):
\[ \sin(17^\circ) = \mu \cos(17^\circ) \]

From this, we can find the coefficient of friction:
\[ \mu = \frac{\sin(17^\circ)}{\cos(17^\circ)} = \tan(17^\circ) \]

### Step 3: Analyze the Forces at \(30^\circ\)
Now, we need to analyze the situation when the angle is increased to \(30^\circ\). The forces acting on the block will now be:
- Perpendicular to the incline: \(mg \cos(30^\circ)\)
- Parallel to the incline: \(mg \sin(30^\circ)\)

The frictional force \(F_r\) will still be:
\[ F_r = \mu R = \mu mg \cos(30^\circ) \]

### Step 4: Write the Equation of Motion
The net force acting on the block when it is at \(30^\circ\) is:
\[ F_{net} = mg \sin(30^\circ) - F_r \]
Substituting for \(F_r\):
\[ F_{net} = mg \sin(30^\circ) - \mu mg \cos(30^\circ) \]

Using \(a = \frac{F_{net}}{m}\):
\[ a = g \sin(30^\circ) - \mu g \cos(30^\circ) \]

### Step 5: Substitute Known Values
We know:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
- \(\mu = \tan(17^\circ) \approx 0.305\)

Substituting these values:
\[ a = g \left(\frac{1}{2}\right) - \left(0.305\right) g \left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Simplify the Expression
Let’s calculate:
\[ a = g \left(\frac{1}{2} - 0.305 \cdot \frac{\sqrt{3}}{2}\right) \]

Calculating \(0.305 \cdot \frac{\sqrt{3}}{2} \approx 0.305 \cdot 0.866 \approx 0.264\):
\[ a = g \left(\frac{1}{2} - 0.264\right) = g \left(0.236\right) \]

Assuming \(g \approx 10 \, \text{m/s}^2\):
\[ a \approx 10 \cdot 0.236 = 2.36 \, \text{m/s}^2 \]

### Final Answer
Thus, the acceleration of the block when the incline is increased to \(30^\circ\) is approximately \(2.36 \, \text{m/s}^2\).

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