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Question #43505

Difficulty: 8/10 | 🟢 Free

The Question

A wedge has an inclined surface that makes an angle θ=37∘\theta = 37^\circ with the horizontal. The inclined surface slopes upward to the right. The wedge accelerates horizontally to the right with a constant acceleration a=5 m/s2a = 5 \text{ m/s}^2. A block of mass m=2 kgm = 2 \text{ kg} is placed on the inclined surface. The coefficient of kinetic friction between the block and the wedge is μk=0.5\mu_k = 0.5. A motor fixed to the wedge pulls the block up the incline using a light string parallel to the incline. The block moves up the incline with a constant relative speed v0=5 m/sv_0 = 5 \text{ m/s} with respect to the wedge. At the instant the wedge has a velocity V=10 m/sV = 10 \text{ m/s} to the right, what is the power delivered by the motor to the block as observed in the ground frame? (Take g=10 m/s2g = 10 \text{ m/s}^2, sin⁡37∘=0.6\sin 37^\circ = 0.6, cos⁡37∘=0.8\cos 37^\circ = 0.8)

Options

  • A.

    325 W325 \text{ W}

  • B.

    125 W125 \text{ W}

  • C.

    140 W140 \text{ W}

  • D.

    403 W403 \text{ W}

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

Power-Velocity Relationship|Kinetic Friction Transition|Pseudo Force Directionality

Why This Question Is Tricky

Common Mistakes

This question tests the frame-dependence of mechanical power and the correct resolution of pseudo-forces.

A common trap is assuming power is frame-invariant, leading students to calculate it in the wedge frame (P=Fv0=125 WP = F v_0 = 125 \text{ W}).

Another lethal misconception is equating the motor's power directly to the rate of change of kinetic energy in the ground frame (140 W140 \text{ W}), ignoring the frame-dependent work done by non-conservative forces (friction) and constraint forces (normal force) which are non-zero when the surface itself is moving.

Finally, students often memorize the normal force on an accelerating wedge as N=m(gcos⁡θ+asin⁡θ)N = m(g\cos\theta + a\sin\theta) without analyzing the geometry, incorrectly adding the pseudo-force component instead of subtracting it, which yields 403 W403 \text{ W}.

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