Question:
Which of the following is true about the coefficient of static friction \((\mu_s)\) and the coefficient of kinetic friction \((\mu_k)\)?
Which of the following is true about the coefficient of static friction \((\mu_s)\) and the coefficient of kinetic friction \((\mu_k)\)?
Show Hint
Always remember:
\[
\mu_s\gt \mu_k.
\]
It is generally harder to start the motion of an object than to keep it moving, which is why the coefficient of static friction is greater than the coefficient of kinetic friction.
Updated On: Jun 26, 2026
- \(\mu_s\) is always equal to \(\mu_k\)
- \(\mu_s\) is always greater than \(\mu_k\)
- \(\mu_s\) is always less than \(\mu_k\)
- Depending upon applications, \(\mu_s\) can be greater, less or equal to \(\mu_k\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand static friction.
Static friction acts between two surfaces when there is no relative motion between them.
The maximum value of static friction is given by \[ f_{s,\max}=\mu_s N, \] where \[ \mu_s \] is the coefficient of static friction and \(N\) is the normal reaction.
Static friction adjusts itself from zero up to its maximum value to prevent motion.
Step 2: Understand kinetic friction.
Once the body starts moving, kinetic friction acts between the surfaces.
Its magnitude is \[ f_k=\mu_k N, \] where \[ \mu_k \] is the coefficient of kinetic friction.
Kinetic friction remains approximately constant during motion.
Step 3: Compare \(\mu_s\) and \(\mu_k\).
Experimentally, it is observed that a larger force is required to start the motion of a body than to keep it moving.
This means \[ f_{s,\max}\gt f_k. \] Since both frictional forces are proportional to the same normal reaction \(N\), \[ \mu_s N\gt \mu_k N. \] Dividing by \(N\), \[ \mu_s\gt \mu_k. \] Thus, the coefficient of static friction is always greater than the coefficient of kinetic friction.
Step 4: Physical explanation.
Before motion starts, the microscopic irregularities of the two surfaces interlock strongly.
More force is required to break these interlocking contacts and initiate motion.
Once the body begins to move, these interlocking contacts do not remain engaged for long durations, reducing the frictional force.
Hence, \[ \mu_k\lt \mu_s. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\mu_s\gt \mu_k} \] Hence, the correct option is \[ \boxed{(2)} \]
Static friction acts between two surfaces when there is no relative motion between them.
The maximum value of static friction is given by \[ f_{s,\max}=\mu_s N, \] where \[ \mu_s \] is the coefficient of static friction and \(N\) is the normal reaction.
Static friction adjusts itself from zero up to its maximum value to prevent motion.
Step 2: Understand kinetic friction.
Once the body starts moving, kinetic friction acts between the surfaces.
Its magnitude is \[ f_k=\mu_k N, \] where \[ \mu_k \] is the coefficient of kinetic friction.
Kinetic friction remains approximately constant during motion.
Step 3: Compare \(\mu_s\) and \(\mu_k\).
Experimentally, it is observed that a larger force is required to start the motion of a body than to keep it moving.
This means \[ f_{s,\max}\gt f_k. \] Since both frictional forces are proportional to the same normal reaction \(N\), \[ \mu_s N\gt \mu_k N. \] Dividing by \(N\), \[ \mu_s\gt \mu_k. \] Thus, the coefficient of static friction is always greater than the coefficient of kinetic friction.
Step 4: Physical explanation.
Before motion starts, the microscopic irregularities of the two surfaces interlock strongly.
More force is required to break these interlocking contacts and initiate motion.
Once the body begins to move, these interlocking contacts do not remain engaged for long durations, reducing the frictional force.
Hence, \[ \mu_k\lt \mu_s. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\mu_s\gt \mu_k} \] Hence, the correct option is \[ \boxed{(2)} \]
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