Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The coefficient of static friction, $\mu_{s}^{2}$ between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. g = 10 m/s^2
Text Solution
Verified by ExpertsThe correct answer is:
D
$\mu_{s} = \frac{m_{B}}{m_{A}} \Rightarrow 0.2 = \frac{m_{B}}{2} \Rightarrow m_{B} = 0.4 \text{ kg}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The coefficient of friction $p_u$ and the angle of friction $\lambda$ are related as
A force of 98 N is required to just start moving a body of mass 100 kg over ice. The coefficient of…
A block weighs W is held against a vertical wall by applying a horizontal force F. The minimum valu…
The maximum static frictional force is
Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a hor…
In the figure shown, a block of weight 10 N resting on a horizontal surface. The coefficient of sta…