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Question #42433

Difficulty: 8/10 | 🟢 Free

The Question

A small block of mass $m$ is placed on a rough articulating wedge. The wedge's angle of inclination $\theta$ with the horizontal is dynamically adjusted over time $t$ such that $\theta(t) = \frac{\pi}{2T^2} t^2$ , where $T$ is a given positive time constant. The block is kept stationary relative to the wedge by a string parallel to the incline. Which of the following represents the time-averaged normal force exerted by the wedge on the block from $t = 0$ to $t = T$ ? (Assume $g$ is the acceleration due to gravity and let $C = \int_0^1 \cos\left(\frac{\pi}{2}x^2\right) dx$ )

Options

A.
$mg C$
B.
$mg \cos\left(\frac{\pi}{8}\right)$
C.
$\frac{\sqrt{3}}{2} mg$
D.
$\frac{2}{\pi} mg$

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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

Normal Force at Angles|Average vs. Instantaneous Calculus|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definition of time-averaged quantities in non-linear dynamical systems.

A common trap for students is conflating the average of a function with the function of an average.

Specifically, students might incorrectly calculate the normal force at the average time $t = T/2$ (yielding $mg \cos(\pi/8)$ ), at the time-averaged angle $\langle \theta \rangle = \pi/6$ (yielding $\frac{\sqrt{3}}{2} mg$ ), or compute the average over the angular displacement instead of time (yielding $\frac{2}{\pi} mg$ ).

By forcing the explicit evaluation of $\frac{1}{\Delta t} \int N(t) dt$ , the problem penalizes algebraic shortcuts and reinforces the necessity of integral calculus when dealing with time-varying forces.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Normal Force at Angles
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free