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Chapter 4: Problem 468

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

Short Answer

Expert verified
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 by step solution

01

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
02

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.
03

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.
04

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.
05

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).

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