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Question #42441

Difficulty: 8/10 | 🟢 Free

The Question

A wedge has a rough inclined surface that makes an angle $\theta = 53^\circ$ with the horizontal, sloping downwards from left to right. A block of mass $m$ is placed on the incline. The coefficient of kinetic friction between the block and the incline is $\mu_k = 0.25$ . The wedge is given a constant horizontal acceleration $a = 5 \text{ m/s}^2$ to the right. Assuming the block is released from rest relative to the wedge, what is the magnitude of the block's acceleration relative to the wedge? (Take $g = 10 \text{ m/s}^2$ , $\sin 53^\circ = 0.8$ , $\cos 53^\circ = 0.6$ )

Options

A.
2.5 \text{ m/s}^2
B.
3.5 \text{ m/s}^2
C.
6.5 \text{ m/s}^2
D.
8.5 \text{ m/s}^2

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

Friction on Inclined Planes|Frame of Reference Dependency|Newton’s Third Law Pairing

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply Newton's laws in a non-inertial reference frame.

The primary trap lies in the calculation of the normal force.

Students frequently memorize $N = mg \cos\theta$ for inclined planes and fail to account for the component of the horizontal pseudo force ( $ma \sin\theta$ ) that acts perpendicular to the surface.

This omission leads to an incorrect, lower normal force ( $N = 6m$ ), which subsequently underestimates the kinetic friction and yields an erroneously high relative acceleration of $3.5 \text{ m/s}^2$ (Option B).

Other traps include ignoring the pseudo force entirely or misassigning its direction along the incline.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Friction on Inclined Planes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free