Laws Of Motion Question 55
Question: A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k. $ {{\mu } _{s}} $ is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q is[IIT-JEE (Screening) 2004]
Options:
A) $ kA $
B) $ \frac{kA}{2} $
C) Zero
D) $ {{\mu } _{s}}mg $
Show Answer
Answer:
Correct Answer: B
Solution:
When two blocks performs simple harmonic motion together then at the extreme position ( at amplitude =A) Restoring force $ F=KA=2ma $
therefore $ a=\frac{KA}{2m} $
There will be no relative motion between P and Q if pseudo force on block P is less than or just equal to limiting friction between P and Q. i.e. $ m\ ( \frac{KA}{2m} )= $ Limiting friction $ \therefore $ Maximum friction $ =\frac{KA}{2} $
