Question

Which of the following is true about the coefficient of static friction (\(\mu_s\)) and the coefficient of kinetic friction (\(\mu_k\))

• A. \(\mu_s\) is always equal to \(\mu_k\)
  % Telugu: \(\mu_s\) ఎల్లప్పుడూ \(\mu_k\)
• B. \(\mu_s\) is always greater than \(\mu_k\)
  % Telugu: \(\mu_s\) ఎల్లప్పుడూ \(\mu_k\) కన్నా ఎక్కువ
• C. \(\mu_s\) is always less than \(\mu_k\)
  % Telugu: \(\mu_s\) ఎల్లప్పుడూ \(\mu_k\) కన్నా తక్కువ
• D. Depending upon applications, \(\mu_s\) can be greater, less or equal to \(\mu_k\) % Telugu: అనువర్తనాలను బట్టి, \(\mu_s\) విలువ \(\mu_k\) కన్నా తక్కువ, ఎక్కువ లేక సమానము

Hint:

\textbf{Static Friction (\(f_s\)):} Opposes impending motion. \(0 \le f_s \le f_{s,max} = \mu_s N\).
\textbf{Kinetic Friction (\(f_k\)):} Opposes motion when surfaces are sliding. \(f_k = \mu_k N\).
Generally, for most surfaces, it's harder to start motion than to maintain it, which implies \(f_{s,max}>f_k\), and thus \(\mu_s>\mu_k\).
In some simplified models or for specific materials, \(\mu_s\) may be taken as approximately equal to \(\mu_k\), but the distinct difference is common.

Answer 1

The Correct Option is B

The coefficient of static friction (\(\mu_s\)) relates to the maximum force of static friction that can exist between two surfaces before motion begins. Static friction \(f_s \le \mu_s N\), where N is the normal force. The coefficient of kinetic friction (\(\mu_k\)) relates to the force of kinetic (or dynamic) friction that acts between two surfaces when they are sliding relative to each other. Kinetic friction \(f_k = \mu_k N\). General observations and principles: \begin{itemize} \item It typically requires more force to start an object moving from rest (overcoming maximum static friction) than to keep it moving at a constant velocity once it has started (overcoming kinetic friction). \item This implies that the maximum static friction is generally greater than the kinetic friction. \item Since \(f_{s,max} = \mu_s N\) and \(f_k = \mu_k N\), if \(f_{s,max}>f_k\), then \(\mu_s N>\mu_k N\), which means \(\mu_s>\mu_k\). \item In some idealized cases or for some specific material pairs, \(\mu_s\) might be very close to or equal to \(\mu_k\), but for most common surfaces, \(\mu_s\) is distinctly greater than \(\mu_k\). The statement "always greater than" is a strong one, but generally \(\mu_s \ge \mu_k\) is true, with strict inequality for most cases. \end{itemize} Let's analyze the options: % Option (a) "\(\mu_s\) is always equal to \(\mu_k\)": Generally false for most materials. % Option (b) "\(\mu_s\) is always greater than \(\mu_k\)": This is the most common and widely accepted relationship in introductory physics and engineering for typical surfaces. While equality can occur in some models or for very specific conditions, the general rule is \(\mu_s>\mu_k\). % Option (c) "\(\mu_s\) is always less than \(\mu_k\)": False. This would mean it's harder to keep an object moving than to start it. % Option (d) "Depending upon applications, \(\mu_s\) can be greater, less or equal to \(\mu_k\)": While variations exist, the general rule for common dry friction is that static friction can be greater than kinetic friction. The "less than" part is not standard. Given the standard understanding, \(\mu_s\) is typically greater than \(\mu_k\). Option (b) reflects this general rule. The phrase "always greater" is very strong. A more precise statement is often \(\mu_s \ge \mu_k\), where equality is rare for real surfaces. However, among the choices, (b) is the best representation of the typical relationship. \[ \boxed{\mu_s \text{ is always greater than } \mu_k} \]