Get Answers to all your Questions

header-bg qa

In the arrangement shown in the figure M_{A}=M_{B}=2\; kg. All strings and pulleys are ideal and friction coefficient between blocks A and B is 0.6. Maximum value of horizontal force F that can be applied so that block A does not slip over the block B will be (consider horizontal surface smooth)

Option: 1

8\; N

Option: 2

12\; N

Option: 3

24\; N


Option: 4

32\; N

Answers (1)


Answer (3)

Over the block will be (consider horizontal surface smooth)

Maximum valueof friction is f_{max}=\mu m_{A}g

f_{max}=0.6 \times2 \times10=12 \; N

Block B moves due to friction. Therefore

a_{max}=\frac{f_{max}}{m_{B}}=\frac{12}{2}=6\; m/s^{2}

Hence, F_{max}=\left ( m_{A}+m_{B} \right )a_{max}=4\times6=24\; N

Posted by


View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks