Problem 159 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(m\) is placed on a surface with a vertical cross-section given by \(y=\frac{x^{3}}{6} .\) If the co-efficient of friction is \(0.5\), the maximum height above the ground at which the block can be placed without slipping is: (A) \(\frac{1}{6} m\) (B) \(\frac{2}{3} m\) (C) \(\frac{1}{3} m\) (D) \(\frac{1}{2} m\)

Step-by-Step Solution

Answer

The maximum height above the ground at which the block can be placed without slipping is: (C) \(\frac{1}{3}m\)

Step 1: Write the equation for the curve and find its slope

The equation for the curve is given by: \[y = \frac{x^3}{6}\] To find the slope of the curve, we will differentiate this equation with respect to x: \[\frac{dy}{dx} = \frac{d}{dx}(\frac{x^3}{6})\] The slope of the curve is: \[\frac{dy}{dx} = \frac{1}{2}x^2\]

Step 2: Compute the angle of inclination

We can find the angle θ between the curve and the horizontal axis using the arctangent function. \[tan(θ) = \frac{dy}{dx}\] So, \[θ = arctan(\frac{dy}{dx})\] \[θ = arctan(\frac{1}{2}x^2)\]

Step 3: Apply equilibrium condition

To find the maximum height at which the block can be placed without slipping, we'll use the equilibrium condition: \[f = mg \times sin(θ)\] Where, f = frictional force m = mass of the block g = acceleration due to gravity The frictional force can be given as: \[f = μN\] Where, μ is the coefficient of friction N = Normal reaction force, which can be given as \(mg \times cos(θ)\) Equating these forces, we get: \[μN = mg \times sin(θ)\]

Step 4: Solve for x and find height (y) without slipping

From the above equation, \[μmg \times cos(θ) = mg \times sin(θ)\] Divide both sides by mg: \[μcos(θ) = sin(θ)\] \[cos(θ) = \frac{sin(θ)}{μ}\] Now we know \(tan(θ) = \frac{sin(θ)}{cos(θ)}\), so replacing cos(θ) in the above equation: \[tan(θ) = \frac{sin(θ)}{\frac{sin(θ)}{μ}}\] Simplify the equation: \[tan(θ) = μ\] Now plug in the value of the coefficient of friction μ = 0.5: \[tan(θ) = 0.5\] Using the angle of inclination equation we derived before: \[\frac{1}{2}x^2 = 0.5\] \[x^2 = 2\] \[x = \sqrt{2}\] Now we can find the height y, by substituting x = √2 in the equation for the curve: \[y = \frac{(\sqrt{2})^3}{6}\] \[y = \frac{2\sqrt{2}}{6}\] \[y = \frac{1}{3}m\] So the maximum height above the ground at which the block can be placed without slipping is: (C) \(\frac{1}{3}m\)

Key Concepts

Friction

Friction is a force that opposes the relative motion of two surfaces in contact. It arises from the interactions at the microscopic level between the surfaces, such as the bumps and grooves that may not be visible to the naked eye. In physics problems, especially those related to mechanics, friction often plays a crucial role in determining the motion or stability of objects.

For a block resting on an inclined surface, the force of friction keeps it from sliding down the slope. The coefficient of friction (denoted as \( \mu \) in the equations) is a dimensionless scalar value that represents the ratio of the frictional force (\( f \)) to the normal force (\( N \)). The normal force is the component of contact force that is perpendicular to the surface, which, in this case, is due to the weight of the block (\( m \times g \)), where \( m \) is the mass and \( g \) is the gravitational acceleration.

In the given problem, we deal with the maximum height at which a block can be placed without slipping due to friction on a surface. The coefficient of friction, \( \mu = 0.5 \), enables us to determine the necessary angle of inclination for the block to start slipping, which further leads to calculating the maximum height. Understanding how friction interacts with the forces on inclined planes is essential for solving such physics problems.

Equations of Motion

The Equations of Motion are fundamental to understanding the dynamics of particles and objects. These equations describe the relationship between an object's position, velocity, acceleration, and time, and they are essential tools for solving a wide range of problems in physics — from simple linear motion to complex projectile and orbital motions.

In the problem at hand, while equations of motion directly concerned with velocity and acceleration are not used, the underlying principles of dynamics are applied through equilibrium conditions. Equilibrium occurs when an object is in a state of balance, where the sum of forces and moments acting upon it equals zero. In a state of equilibrium, a body will remain at rest or move with constant velocity.

The principle of equilibrium is leveraged to find the point at which the forces of friction and gravitational force balance each other. This ensures that the mass does not move, hence finding the maximum height at which the block can be placed on the curve without slipping. Therefore, a deep understanding of the equations of motion and the principles of equilibrium are necessary for problem-solving in physics, particularly in situations involving frictional forces.

Differentiation

Differentiation is a fundamental concept in calculus, concerned with calculating the rate at which a function changes at any point. In practical terms, it tells us how a quantity, such as velocity or slope, changes in response to changes in another quantity, such as time or position. Differentiation gives us the derivative of a function, which is extensively used in physics to express a variety of rates like acceleration (the rate of change of velocity) or the slope of a curve in a graph.

In the problem provided, differentiation is used to determine the slope of the curve representing the surface on which the block rests. The curve is given by the equation \( y = \frac{x^3}{6} \), and its derivative with respect to \( x \), found using differentiation, reveals how steep the slope is at any point. The derivative \( \frac{dy}{dx} = \frac{1}{2}x^2 \) represents the slope and is used to calculate the angle of inclination of the surface at the point where the block is positioned.

Understanding how to derive and interpret a function's derivative is indispensable when analyzing the motion of objects over varying paths, as it aids in exposing the relationship between forces such as gravity, friction, and the geometry of the path an object moves along. In physics, especially in problems involving motion and forces, differentiation helps reveal these relationships, making it a key tool in solving complex problems.