Discussion
If the maximum acceleration of a moving platform to keep a box of mass 5 kg on it without sliding is \(3ms^{-1}\) , then the static friction between the box and floor of the platform is \((g = 10ms^{-2})\)
Answer: C
The coefficient of static friction is 0.30.
The box accelerates with the platform only because of static friction, which can supply at most \(\mu mg\). So the maximum acceleration without sliding is \(a_{max} = \mu g\).
\[ \mu = \frac{a_{max}}{g} = \frac{3}{10} = 0.30 \]
The mass of 5 kg cancels and is not needed. (The acceleration is printed with the unit m s-1; it is an acceleration of 3 m s-2.) 0.15 and 0.4 do not equal \(a/g\).
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development