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Question #43504

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass m=2 kgm = 2 \text{ kg} moves on a smooth horizontal plane. Its position in polar coordinates (r,θ)(r, \theta) is given by r(t)=2+sin⁡(πt)r(t) = 2 + \sin(\pi t) meters and θ(t)=2t\theta(t) = 2t radians, where tt is the time in seconds. The particle is subjected to a net force F⃗\vec{F} which can be resolved into a radial component FrF_r and a tangential component FtF_t. Let WtW_t be the work done by the tangential force FtF_t and WnetW_{\text{net}} be the work done by the net force F⃗\vec{F} on the particle during the time interval t=0t = 0 to t=0.5 st = 0.5 \text{ s}. Which of the following gives the correct values of (Wt,Wnet)(W_t, W_{\text{net}})?

Options

  • A.

    (20−π2 J,20−π2 J)(20 - \pi^2 \text{ J}, 20 - \pi^2 \text{ J})

  • B.

    (40 J,20−π2 J)(40 \text{ J}, 20 - \pi^2 \text{ J})

  • C.

    (40 J,40 J)(40 \text{ J}, 40 \text{ J})

  • D.

    (20+π2 J,40 J)(20 + \pi^2 \text{ J}, 40 \text{ J})

Check Your Answer

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

Power-Velocity Relationship|Non-Uniform Circular Motion|Principle of Homogeneity Exceptions

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the work done by a tangential force and the net work done on a particle in a system with a varying radius.

A common misconception is assuming that the radial (centripetal) force never does work, leading students to incorrectly equate the net work (change in kinetic energy) entirely to the work done by the tangential force (Wt=ΔKW_t = \Delta K).

By introducing a time-dependent radius, the radial velocity r˙\dot{r} becomes non-zero, meaning the radial force FrF_r does work (Wr=∫FrdrW_r = \int F_r dr).

Students who fall for the trap will calculate ΔK\Delta K and incorrectly select the option where Wt=Wnet=20−π2 JW_t = W_{\text{net}} = 20 - \pi^2 \text{ J}, failing to integrate the tangential power Pt=FtvtP_t = F_t v_t independently.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptPower-Velocity Relationship
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree