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Question #42032

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m = 2 \text{ kg}$ is constrained to move along the $x$ -axis under the influence of a conservative force field. The potential energy of the particle is given by $U(x) = \left(-\frac{1}{3}x^3 + 2x^2 - 3x\right) \text{ J}$ , where $x$ is the position in meters. If the particle is released from rest at the origin ( $x = 0$ ), what is its speed when its acceleration becomes zero for the first time?

Options

A.
$\frac{2}{\sqrt{3}} \text{ m/s}$
B.
$\sqrt{3} \text{ m/s}$
C.
$\sqrt{\frac{3}{2}} \text{ m/s}$
D.
$\frac{2\sqrt{2}}{\sqrt{3}} \text{ m/s}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

The Velocity-Position Identity|Conservative Force Field|Calculus (Integration)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from time-domain dynamics to spatial-domain dynamics in a conservative force field.

A common trap is attempting to use constant acceleration kinematic equations (like $v^2 = u^2 + 2as$ ) by incorrectly assuming the initial acceleration ( $a = 1.5 \text{ m/s}^2$ ) remains constant, which leads directly to the trap answer $\sqrt{3} \text{ m/s}$ .

By providing a non-linear potential energy function, the problem forces the explicit use of the $v \frac{dv}{dx}$ substitution (or equivalently, the work-energy theorem) to correctly integrate the position-dependent acceleration.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	The Velocity-Position Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free