A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k, the two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks?

Question

A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k, the two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks? (A) kA/2 (B) kA (C) μs mg  (D) Zero

Tardigrade Admin · Accepted Answer

Angular frequency of the system, ω = √(k/m+m) = √(k/2m) Maximum acceleration of the system will be ω2 A or (kA/2m) This acceleration to the lower block is provided by friction. Hence , fmax = mamax = m ω2 A = m( ( kA/2m)) = (kA/2)

$k A /2$

kA

$μ_{s} m g$

Zero

Angular frequency of the system, $ω = \frac{k}{m + m} = \frac{k}{2 m}$
Maximum acceleration of the system will be $ω^{2} A or \frac{k A}{2 m}$
This acceleration to the lower block is provided by
friction.
Hence , $f_{ma x} = m a_{ma x} = m ω^{2} A = m (\frac{k A}{2 m}) = \frac{k A}{2}$

kA/2

kA

μs​mg

Zero

Angular frequency of the system, ω=m+mk​​=2mk​​ Maximum acceleration of the system will be ω2Aor2mkA​ This acceleration to the lower block is provided by friction. Hence , fmax​=mamax​=mω2A=m(2mkA​)=2kA​

$k A /2$

$μ_{s} m g$

Angular frequency of the system, $ω = \frac{k}{m + m} = \frac{k}{2 m}$
Maximum acceleration of the system will be $ω^{2} A or \frac{k A}{2 m}$
This acceleration to the lower block is provided by
friction.
Hence , $f_{ma x} = m a_{ma x} = m ω^{2} A = m (\frac{k A}{2 m}) = \frac{k A}{2}$