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Question #43849

Difficulty: 8/10 | 🟢 Free

The Question

A light spring of force constant k=100 N/mk = 100 \text{ N/m} hangs vertically from a fixed support. A small ring of mass m=1.0 kgm = 1.0 \text{ kg} is attached to the lower end of the spring and is threaded on a rough vertical guide rod. A constant horizontal force F=20 NF = 20 \text{ N} presses the ring against the rod. The coefficient of kinetic friction between the ring and the rod is μk=0.20\mu_k = 0.20, and the coefficient of static friction is μs=0.30\mu_s = 0.30. The ring is pulled downwards to a distance of 39 cm39 \text{ cm} from the relaxed position of the spring and released from rest. Taking g=10 m/s2g = 10 \text{ m/s}^2, what is the total distance traveled by the ring before it comes to a permanent stop?

Options

  • A.

    102 cm102 \text{ cm}

  • B.

    104 cm104 \text{ cm}

  • C.

    105.1 cm105.1 \text{ cm}

  • D.

    190 cm190 \text{ cm}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Vertical Oscillations|SHM Definition via Differential Equation|Kinetic Friction Transition

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model a damped oscillator with Coulomb (dry) friction, specifically focusing on the transition from kinetic to static friction at the boundary condition.

A common trap is assuming the motion continues until the energy is completely dissipated continuously (yielding 105.1 cm) or incorrectly using the kinetic friction coefficient for the final stopping condition (yielding 104 cm, as the student would calculate one extra micro-oscillation).

Students must recognize that the system stops discretely when the velocity is zero and the restoring force is less than the maximum static friction, requiring a step-by-step evaluation of the amplitude decay.

Furthermore, the vertical setup requires shifting the coordinate system to the new equilibrium position to correctly apply the symmetric amplitude drop formula.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterOscillations - Class 11 Part 2
ConceptVertical Oscillations
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree