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Question #42051

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $m = 4 \text{ kg}$ is initially at rest on a horizontal floor. At $t = 0$ , a time-varying horizontal force $F(t) = 8t$ (where $t$ is in seconds and $F$ is in Newtons) is applied to the block. Due to a delayed engagement mechanism, the floor is perfectly frictionless for the first $0.5 \text{ s}$ . At exactly $t = 0.5 \text{ s}$ , the friction mechanism engages, and the floor instantly provides a coefficient of static friction $\mu_s = 0.4$ and kinetic friction $\mu_k = 0.2$ . Taking $g = 10 \text{ m/s}^2$ , what is the total distance traveled by the block from $t = 0$ to $t = 1.5 \text{ s}$ ?

Options

A.
$\frac{1}{12} \text{ m}$
B.
$\frac{1}{8} \text{ m}$
C.
$\frac{1}{24} \text{ m}$
D.
0 \text{ m}

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

The Reaction Time Gap|Static Friction Threshold|Recoil and Thrust Forces

Why This Question Is Tricky

Common Mistakes

This question tests the 'non-instantaneous constraint' trap and the transition between kinetic and static friction.

Students often reflexively apply the static friction threshold ( $F_{max} = \mu_s N$ ) from $t=0$ , incorrectly concluding the block remains at rest until $t=2$ s (Option D).

Alternatively, if they correctly identify the initial frictionless drift, they may blindly integrate the velocity function $(t-1)^2$ past $t=1$ s (Option B), failing to realize the block momentarily stops and gets trapped by static friction because the applied force ( $8$ N) is less than the static threshold ( $16$ N).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	The Reaction Time Gap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free