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Question #42468

Difficulty: 8/10 | 🟢 Free

The Question

A wedge of mass $M = 2m$ and inclination angle $\theta = 45^\circ$ rests on a smooth horizontal surface. A block of mass $m$ is placed on the rough inclined surface of the wedge and released from rest. The coefficient of kinetic friction between the block and the wedge is $\mu = 0.5$ . What is the total work done by the friction force on the system (block + wedge) by the time the block descends a vertical height $h$ ?

Options

A.
$-\frac{4}{9} mgh$
B.
$-\frac{1}{2} mgh$
C.
$-\frac{11}{27} mgh$
D.
$-\frac{5}{9} mgh$

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

Wedge Constraint Logic|Dot Product Geometry|Kinetic Friction Transition

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in a multi-body system with non-conservative internal constraint forces.

A common trap is assuming the normal force remains $mg \cos\theta$ as it would on a fixed incline, leading to an incorrect frictional force and a trap answer of $-0.5 mgh$ .

Students must realize that the wedge's acceleration alters the normal force (requiring D'Alembert's principle or pseudo forces).

Furthermore, they must correctly identify that the net work done by kinetic friction on the system depends strictly on the relative displacement between the surfaces, not the individual ground frame displacements.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Wedge Constraint Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free