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Question #43476

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass m=2 kgm = 2 \text{ kg} is moving along the positive x-axis. At t=0t = 0, its velocity is v0=2 m/sv_0 = 2 \text{ m/s}. It experiences a net force in the direction of its motion given by F=kv2F = k v^2, where k=0.5 kg/mk = 0.5 \text{ kg/m} and vv is its instantaneous velocity. What is the average power delivered by this force to the particle during the time interval from t=0t = 0 to t=1 st = 1 \text{ s}?

Options

  • A.

    12 W12 \text{ W}

  • B.

    13.5 W13.5 \text{ W}

  • C.

    15 W15 \text{ W}

  • D.

    18 W18 \text{ W}

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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

Average vs. Instantaneous Calculus|Power-Velocity Relationship|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between instantaneous power (P=FvP = Fv) and average power (Pavg=ΔW/ΔtP_{avg} = \Delta W / \Delta t).

A common student trap is to assume that average power can be calculated by taking the arithmetic mean of the initial and final instantaneous powers (Pavg=(Pi+Pf)/2=18 WP_{avg} = (P_i + P_f)/2 = 18 \text{ W}), or by evaluating the power function at the arithmetic mean of the velocities (P=kvmean3=13.5 WP = k v_{mean}^3 = 13.5 \text{ W}).

These algebraic shortcuts fail because the force and power are non-linear functions of velocity and time.

The rigorous approach requires integrating the equation of motion to find the final state, and then applying the Work-Energy Theorem to find the total work done over the interval before dividing by the time elapsed.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Straight Line - Class 11 Part 1
ConceptAverage vs. Instantaneous Calculus
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree