Laws Of Motion Question 336
Question: The two blocks, $ m=10kg $ and $ M=50kg $ are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to
Options:
A) 100 N
B) 50 N
C) 240 N
D) 180 N
Show Answer
Answer:
Correct Answer: C
Solution:
[c]As m would slip in vertically downward direction, then $ mg=\mu N $
$ \Rightarrow N=\frac{mg}{\mu }=\frac{100}{0.5}=200Newton $
Same normal force would accelerated M, thus $ a _{M}=\frac{200}{50}=4m/s^{2} $ Taking $ m+M $ as system $ F=(m+M)4=240N $
BETA
