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Question #42750

Difficulty: 8/10 | 🟢 Free

The Question

A small block of mass mm and positive charge qq rests on a rough horizontal table. It is constrained to move along a circular groove of radius RR centered at the origin. The coefficient of kinetic friction between the block and the table is μ\mu. A stationary positive point charge QQ is fixed at a height hh directly above the center of the circle. The block is given an initial tangential speed v0v_0. The groove provides only the necessary radial force to maintain the circular path and exerts no friction. Which of the following expressions gives the total distance ss traveled by the block before coming to rest? (Let k=14πϵ0k = \frac{1}{4\pi\epsilon_0})

Options

  • A.

    mv022μ(mg+kQqh(R2+h2)3/2)\frac{m v_0^2}{2 \mu \left( mg + \frac{k Q q h}{(R^2+h^2)^{3/2}} \right)}

  • B.

    v022μg\frac{v_0^2}{2 \mu g}

  • C.

    mv022μ(mg+kQqR(R2+h2)3/2)\frac{m v_0^2}{2 \mu \left( mg + \frac{k Q q R}{(R^2+h^2)^{3/2}} \right)}

  • D.

    ∞\infty

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Equipotential Surface Geometry|Work-Energy Theorem Validity|Work Done by Friction

Why This Question Is Tricky

Common Mistakes

This question tests the validity of the Work-Energy Theorem in the presence of both conservative and non-conservative forces on an equipotential surface.

The primary trap exploits the misconception that motion along an equipotential surface implies zero net work and thus constant kinetic energy (leading to the incorrect choice of infinity).

Students often erroneously conflate ΔU=0\Delta U = 0 with ΔK=0\Delta K = 0, ignoring the path-dependent, dissipative work done by kinetic friction.

Furthermore, a secondary trap exists where students recognize friction but fail to account for the vertical component of the electrostatic force augmenting the normal force, mistakenly assuming the electric field has absolutely no dynamical effect simply because it does no work.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptEquipotential Surface Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree