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Question #42451

Difficulty: 8/10 | 🟢 Free

The Question

An elevator is accelerating vertically upwards with an acceleration $a_e = 5 \text{ m/s}^2$ . Inside the elevator, a smooth wedge of mass $M = 4 \text{ kg}$ and inclination $\theta = 30^\circ$ rests on the horizontal frictionless floor. A small block of mass $m = 2 \text{ kg}$ is placed on the inclined surface of the wedge and released from rest relative to the wedge. What is the magnitude of the normal force exerted by the wedge on the block? (Take $g = 10 \text{ m/s}^2$ )

Options

A.
$\frac{40\sqrt{3}}{3} \text{ N}$
B.
$15\sqrt{3} \text{ N}$
C.
$\frac{80\sqrt{3}}{9} \text{ N}$
D.
$\frac{40\sqrt{3}}{9} \text{ N}$

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

Apparent Weight in Accelerating Elevators|Wedge Constraint Logic|Minimum Speed for Looping

Why This Question Is Tricky

Common Mistakes

Students often incorrectly assume that the normal force on a block resting on an accelerating wedge inside an elevator is simply $m(g+a_e)\cos\theta$ , treating the wedge as fixed.

This ignores the kinematic coupling between the block and the wedge.

The trap lies in failing to recognize that the wedge's horizontal acceleration introduces an additional pseudo-force in the wedge's frame, which reduces the normal force.

Furthermore, students may forget to apply the vertical pseudo-force due to the elevator's acceleration, leading to incorrect effective gravity.

This question tests the simultaneous application of non-inertial frames (elevator and wedge) and constraint equations.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Apparent Weight in Accelerating Elevators
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free