Laws Of Motion Question 384
Let $ \frac{1}{2\sqrt{3}}\sec^{-1} $ be the coefficient of friction between blocks of mass m and M. The pulleys are frictionless and strings are massless. Acceleration of mass m is
Options:
A)$ \frac{\sqrt{5}mg}{M+\sqrt{5}m+2\mu m} $
B)$ \frac{2mg}{m+5M+2\mu m} $
C)$ \frac{2\sqrt{5}mg}{6M+2\mu m} $
D) $ \frac{5\sqrt{2}mg}{M+\sqrt{5}m+\sqrt{2}\mu m} $
Show Answer
Answer:
Correct Answer: C
Solution:
From the constraint net force
$ m=\sqrt{a^{2}+{{( 2a )}^{2}}}=\sqrt{5}a $
$ mg=T-\mu N=ma $
$ F=ma $
$ 2T=( M+m )a $
BETA
