Problem 468 - JEE Physics

Question

The potential energy of \(2 \mathrm{~kg}\) particle, free to move along \(\mathrm{x}\) axis is given by \(\mathrm{U}(\mathrm{X})=\left[\left\\{\mathrm{x}^{4} / 4\right\\}-\left\\{\mathrm{x}^{2} / 2\right\\}\right] \mathrm{J}\). If its mechanical energy is \(2 \mathrm{~J}\), its maximum speed is \(\ldots \mathrm{m} / \mathrm{s}\) (A) \((3 / 2)\) (B) \(\sqrt{2}\) (C) \((1 / \sqrt{2})\) (D) 2

Step-by-Step Solution

Answer

The maximum speed of the 2 kg particle is \( \boxed{ \sqrt{ \frac{9}{2} } } \) m/s, which is approximately equal to (3/2) m/s. The correct answer is (A).

Step 1: Write down the known quantities

We are given the following information: - Mass of the particle, m = 2 kg - Potential energy function, U(x) = (x^4/4) - (x^2/2) J - Total mechanical energy, E = 2 J

Step 2: Calculate the kinetic energy at the maximum speed

According to conservation of mechanical energy, at any point, the sum of the kinetic energy and the potential energy is constant. This is expressed by the equation: E = K + U We are given E and U(x), so we can solve for K: K = E - U(x) The maximum speed occurs when the kinetic energy is at its highest value, which means the potential energy is at the lowest possible value, given the constant total energy. To find this value, we can use calculus to find the critical points of the potential energy function, U'(x)=0.

Step 3: Find the critical points of the potential energy function

Differentiate U(x) with respect to x: U'(x) = d( (x^4/4) - (x^2/2) )/dx U'(x) = x^3 - x To find the critical points, set U'(x) = 0: x^3 - x = 0 x(x^2 - 1) = 0 x(x - 1)(x + 1) = 0 There are three critical points: x = 0, x = 1, and x = -1.

Step 4: Find the minimum potential energy and its corresponding x value

Calculate the potential energy at the critical points: U(0) = 0 U(1) = (1/4) - (1/2) = -1/4 J U(-1) = (-1/4) - (1/2) = -1/4 J The minimum potential energy is -1/4 J at both x = 1 and x = -1.

Step 5: Calculate the maximum kinetic energy and speed

Now that we have found the minimum potential energy, we can find the maximum kinetic energy: K(max) = E - U(min) K(max) = 2 - (-1/4) = 9/4 J Using the formula for kinetic energy, K = (1/2)mv^2, we can solve for the maximum speed, v: (1/2)(2 kg)v^2 = (9/4) J v^2 = (9/2) m^2/s^2 v = sqrt(9/2) m/s So the maximum speed is \( \boxed{ \sqrt{ \frac{9}{2} } } \) m/s, which is approximately equal to (3/2) m/s. The correct answer is (A).

Key Concepts

Potential Energy

Potential energy is the energy stored in an object due to its position or condition. In this exercise, the potential energy function is given as\[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \]This equation describes how the potential energy changes as the particle moves along the x-axis. Potential energy can vary with position, and here we see it is dependent on the power of x.- If the x value increases, the potential energy function involves terms of higher powers of x.- It plays a crucial role in the conservation of energy, as it contributes to the total mechanical energy along with kinetic energy.Understanding potential energy can help predict how systems behave, notably showing where energy is stored and from where it can be released during movement or interaction.

Kinetic Energy

Kinetic energy is the energy of motion. It is calculated using the formula:\[ K = \frac{1}{2}mv^2 \]- **m** stands for mass (in this problem, it is 2 kg).- **v** is the velocity or speed of the object.When potential energy is minimum, kinetic energy reaches its maximum value. In this context:- The maximum kinetic energy was found by subtracting the minimum potential energy from the total mechanical energy.- This allowed for calculating the maximum speed the particle can achieve upon reaching its peak kinetic energy.Kinetic energy is dynamic and changes with the velocity of the particle. This exercise highlights its importance in understanding how energy transitions between forms as conditions change.

Critical Points

Critical points are values of **x** where the derivative of a function is zero or undefined. In our exercise, we examined the critical points of the potential energy derivative to find where minimum potential energy occurs.- We differentiated the potential energy function: \[ U'(x) = x^3 - x \]- Setting the derivative equal to zero, we solved for x to get critical points: - Solving gives values of x: \[ x = 0, x = 1, x = -1 \]- By evaluating the potential energy at these points, we determine where the energy is minimized.Understanding critical points is essential in calculus as they help locate where functions reach maximum or minimum values, providing insights into system behaviors and optimizations.

Differentiation

Differentiation is a process in calculus that helps in finding the rate at which a function changes. In the exercise, differentiation was used to find the critical points of the potential energy function, helping determine the point of minimum potential energy.- Differentiating the potential energy function, \[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \], results in the derivative \[ U'(x) = x^3 - x \].- Setting the derivative equal to zero, \[ x^3 - x = 0 \], allowed for identifying critical points where potential energy could be extremized.Differentiation provides the tools necessary to find how functions change and is particularly useful in physics for understanding how different quantities interact and evolve over time. This method is crucial in finding insights into mechanical systems and their energy dynamics.