An inclined plane is bent in such a way that the vertical cross-section is given by y=`(x^2)/(4)` where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction `mu` = 0.5, the maximum height in cm at which a stationary block will not slip downward is______cm.

Answer

To solve the problem, we need to determine the maximum height at which a stationary block will not slip down an inclined plane defined by the equation \( y = \frac{x^2}{4} \). Given that the coefficient of friction \( \mu = 0.5 \), we will follow these steps:

### Step 1: Understand the Geometry of the Inclined Plane
The equation \( y = \frac{x^2}{4} \) describes a parabola that opens upwards. The height \( y \) at any horizontal distance \( x \) is given by this equation.

### Step 2: Find the Slope of the Inclined Plane
To find the slope of the inclined plane at any point, we need to differentiate the equation with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{4}\right) = \frac{2x}{4} = \frac{x}{2} \]

### Step 3: Determine the Angle of Inclination
The slope \( \frac{dy}{dx} \) gives us \( \tan \theta \), where \( \theta \) is the angle of inclination of the plane:

\[ \tan \theta = \frac{x}{2} \]

### Step 4: Apply the Condition for No Slipping
For the block to remain stationary, the angle of inclination must satisfy the condition:

\[ \tan \theta = \mu \]

Given \( \mu = 0.5 \), we have:

\[ \tan \theta = 0.5 \]

### Step 5: Set the Two Expressions for \( \tan \theta \) Equal
Now, we equate the two expressions for \( \tan \theta \):

\[ \frac{x}{2} = 0.5 \]

### Step 6: Solve for \( x \)
Multiplying both sides by 2 gives:

\[ x = 1 \]

### Step 7: Substitute \( x \) Back to Find \( y \)
Now that we have \( x = 1 \), we can substitute this value back into the equation for \( y \):

\[ y = \frac{x^2}{4} = \frac{1^2}{4} = \frac{1}{4} \]

### Step 8: Convert Height to Centimeters
Since the height \( y \) is in meters, we convert it to centimeters:

\[ y = \frac{1}{4} \text{ m} = 0.25 \text{ m} = 25 \text{ cm} \]

### Final Answer
The maximum height at which a stationary block will not slip downward is **25 cm**.

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