Laws of Motion - Result Question 29
29. The coefficient of static friction, $\mu_s$, 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}) $
(a) $0.4 kg$
(b) $2.0 kg$
(c) $4.0 kg$
(d) $0.2 kg$
Show Answer
Answer:
Correct Answer: 29. (a)
Solution:
- (a) Condition for the max value of mass of block B so that two blocks do not move is,
$m_B g=\mu_s m_A g$
$\Rightarrow m_B=\mu_s m_A$
or, $m_B=0.2 \times 2=0.4 kg$
