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Question #42302

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $m$ is released from rest on a wedge of mass $M$ and inclination angle $\theta$ . The wedge rests on a smooth horizontal surface. The coefficient of kinetic friction between the block and the wedge varies as $\mu = kx$ , where $x$ is the distance the block has slid relative to the wedge down the incline. Which of the following expressions correctly represents the normal force $N$ exerted by the wedge on the block as a function of $x$ ?

Options

A.
$\frac{M mg \cos\theta}{M + m \sin^2\theta - m kx \sin\theta \cos\theta}$
B.
$\frac{M mg \cos\theta}{M + m \sin^2\theta}$
C.
$mg \cos\theta$
D.
$\frac{M mg \cos\theta}{M - m \sin^2\theta + m kx \sin\theta \cos\theta}$

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|Work Done by Friction|FBD Isolation

Why This Question Is Tricky

Common Mistakes

This question targets the lethal misconception that the normal force on an incline is a static constant ( $N = mg \cos\theta$ ) or independent of non-conservative forces.

By introducing a position-dependent friction coefficient ( $\mu = kx$ ), the friction force alters the wedge's horizontal acceleration dynamically.

Because the wedge's acceleration dictates the pseudo-force (or absolute acceleration component) perpendicular to the incline, the normal force itself becomes a function of position.

Students who blindly apply standard work-energy formulas or uncoupled free-body diagrams will fail to capture the $kx$ dependence in the normal force, testing their ability to resolve coupled non-inertial reference frames.

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