Question

A particle of mass = 1 kg lying on x -axis experience a force given by law F = x(3x - 2) Newton, where x is the x -coordinate of the particle in meters. What is the minimum speed to be imparted to the particle placed at x = 4 meters such that it reaches the origin.

• A. $\sqrt{\frac{2600}{27}}$ m/s
• B. $\sqrt{\frac{2300}{27}}$ m/s
• C. $\sqrt{\frac{2500}{27}}$
• D. $\sqrt{\frac{2500}{23}}$ m/s

Answer 1

Answer: (A)

$\sqrt{\frac{2600}{27}}$ m/s

Answer 2

The problem asks for the minimum speed to be imparted to a particle of mass m = 1 kg at x = 4 meters such that it reaches the origin (x = 0). The force acting on the particle is given by F = x(3x - 2) Newton.

1. Determine the potential energy function U(x): The force F is related to the potential energy U by the equation F = -dU/dx. So, dU = -F dx. Given F = x(3x² - 2x). Integrating to find U(x): $U(x) = -\int (3x^2 - 2x) dx$ $U(x) = -(3\frac{x^3}{3} - 2\frac{x^2}{2}) + C$ $U(x) = -(x^3 - x^2) + C$ We can set the integration constant C = 0, as only potential energy differences are physically significant. $U(x) = x^2 - x^3$

2. Analyze the potential energy landscape: We need to find the potential energy at the initial position (x = 4), the final position (x = 0), and any critical points (local maxima or minima) between them. Critical points occur where F(x) = -dU/dx = 0. $F(x) = x(3x - 2) = 0$ This gives two critical points: x = 0 and x = 2/3.

   Let's evaluate the potential energy at these points:

   • At x = 0 (origin): $U(0) = 0^2 - 0^3 = 0$
   • At x = 2/3: $U(2/3) = (2/3)^2 - (2/3)^3 = \frac{4}{9} - \frac{8}{27} = \frac{12}{27} - \frac{8}{27} = \frac{4}{27}$
   • At x = 4 (initial position): $U(4) = 4^2 - 4^3 = 16 - 64 = -48$

   Plotting these values or considering the second derivative of U(x) (or first derivative of F(x)) helps determine the nature of critical points. F(x) = 3x² - 2x. dF/dx = 6x - 2. At x = 0, dF/dx = -2 < 0, which means U(0) is a local minimum (stable equilibrium). At x = 2/3, dF/dx = 6(2/3) - 2 = 4 - 2 = 2 > 0, which means U(2/3) is a local maximum (unstable equilibrium).

   The particle starts at x = 4 with U(4) = -48 J. It needs to reach x = 0 with U(0) = 0 J. On the path from x = 4 to x = 0, the particle must pass through the potential energy barrier at x = 2/3, where U(2/3) = 4/27 J. Since U(2/3) = 4/27 J is greater than both U(4) = -48 J and U(0) = 0 J, this point represents the highest potential energy the particle must overcome to reach the origin.

3. Apply the principle of conservation of mechanical energy: For the particle to reach the origin, its total mechanical energy (E = K + U) must be at least equal to the maximum potential energy it encounters on its path. For minimum initial speed, the kinetic energy at the peak of the barrier (x = 2/3) should be zero. Let v be the minimum speed imparted at x = 4. Initial energy E_initial = K_initial + U_initial = (1/2)mv² + U(4) Energy at the barrier E_barrier = K_barrier + U_barrier = 0 + U(2/3) (for minimum speed)

   Setting E_initial = E_barrier: $(1/2)mv^2 + U(4) = U(2/3)$ Given m = 1 kg: $(1/2)(1)v^2 + (-48) = 4/27$ $(1/2)v^2 = 4/27 + 48$ To sum the terms on the right: $(1/2)v^2 = \frac{4 + 48 \times 27}{27}$ Calculate 48 × 27: $48 \times 27 = 48 \times (30 - 3) = 1440 - 144 = 1296$ Substitute this value back: $(1/2)v^2 = \frac{4 + 1296}{27}$ $(1/2)v^2 = \frac{1300}{27}$ $v^2 = \frac{2600}{27}$ $v = \sqrt{\frac{2600}{27}} \text{ m/s}$

This is the minimum speed required for the particle to cross the potential energy barrier at x = 2/3 and subsequently reach the origin.