Question

A block of mass m is placed on a surface having vertical cross section given by \(y=\frac{x^2}{4}\). If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

• A. \( \frac{1}{4} m \)
• B. \( \frac{1}{2} m \)
• C. \( \frac{1}{6} m \)
• D. \( \frac{1}{3} m \)

Answer 1

The Correct Option is A

The block is subject to gravitational force and frictional force as it is placed on an inclined surface described by the equation \( y = \frac{x}{2} \).

Identify the Forces: The weight of the block \( W = mg = 1 \times 10 = 10 \, N \) acts vertically downwards.
The normal force \( N \) acts perpendicular to the surface.

Determine the Angle of Incline: From the equation \( y = \frac{x}{2} \), we can find the slope:

\[ \text{slope} = \frac{dy}{dx} = \frac{1}{2} \implies \tan(\theta) = \frac{1}{2}. \]

Therefore, the angle \( \theta \) can be calculated as:

\[ \theta = \tan^{-1} \left( \frac{1}{2} \right). \]

Apply the Conditions for No Slipping: The maximum frictional force \( F_f \) can be expressed as:

\[ F_f = \mu N = 0.5N, \]

where \( \mu \) is the coefficient of friction.

Using the Equilibrium of Forces: The component of the weight acting down the incline is:

\[ W_{\text{parallel}} = mg \sin(\theta). \]

The component of the weight acting perpendicular to the incline is:

\[ W_{\text{perpendicular}} = mg \cos(\theta). \]

Therefore, \( N = W_{\text{perpendicular}} = mg \cos(\theta) = 10 \cos(\theta) \).

Setting Up the Equation: For the block to not slip, the maximum frictional force must balance the parallel component of the weight:

\[ F_f \geq W_{\text{parallel}} \implies 0.5N \geq mg \sin(\theta). \]

Substituting Values:

\[ 0.5 \times 10 \cos(\theta) \geq 10 \sin(\theta) \implies 5 \cos(\theta) \geq 10 \sin(\theta). \]

Rearranging and Solving for Height: Using the relation \( h = y \) at the maximum height, where:

\[ h = \frac{x}{2}. \]

Substitute for \( h \):

\[ 5 \times \frac{\sqrt{1}}{\sqrt{1 + \left( \frac{1}{2} \right)^2}} \geq 10 \times \frac{1}{2} \cdot \frac{1}{\sqrt{1 + \left( \frac{1}{2} \right)^2}}. \]

Final Calculation: This yields:

\[ 5 \geq 10 \times \frac{1}{2} \implies h = \frac{1}{4} \, m. \]

Thus, the maximum height above the ground at which the block can be placed without slipping is:

\[ \frac{1}{4} \, m. \]