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Question #42036

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $m$ is projected with an initial velocity $u$ along a rough horizontal surface having a coefficient of kinetic friction $\mu$ . In addition to friction, the block experiences an aerodynamic drag force $F_d = -kv$ , where $k$ is a positive constant and $v$ is the instantaneous velocity of the block. Acceleration due to gravity is $g$ . What is the total distance traveled by the block before it comes to rest?

Options

A.
$\frac{mu}{k} - \frac{\mu m^2 g}{k^2} \ln\left(1 + \frac{ku}{\mu mg}\right)$
B.
$\frac{mu}{k} + \frac{\mu m^2 g}{k^2} \ln\left(1 + \frac{ku}{\mu mg}\right)$
C.
$\frac{m^2 u}{\mu k} - \frac{m g}{k} \ln\left(1 + \frac{ku}{\mu mg}\right)$
D.
$\frac{u^2}{2\mu g} - \frac{m}{k} \ln\left(1 + \frac{ku}{\mu mg}\right)$

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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

Stopping Distance Paradox|Viscosity and Stokes' Law|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the transition from constant acceleration kinematics to calculus-based dynamics.

A common trap is assuming the stopping distance can be found using $v^2 = u^2 + 2as$ by plugging in an 'average' acceleration or treating the drag force as a simple work-energy term without proper integration.

Students must recognize that when force depends on velocity, they must use the differential form $a = v(dv/dx)$ and perform a logarithmic integration, punishing those who rely on standard algebraic equations of motion.

Quick Facts About This Question

Question Information

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