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PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

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A block of mass $M$ slides on a frictionless surface with an initial speed of $v_{0}$ . On top of the block, there is a small block of mass $m$ . The coefficient of friction between the blocks is $μ$ . The system of blocks encounters an ideal spring with force constant $k$ . The maximum value of $k$ for which the blocks don't slip relative to each other is

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Answer:

Solution:

$\frac{1}{2} (M + m) (v_{0})^{2} = \frac{1}{2} k x^{2}$

$x = v_{0} \sqrt{\frac{M + m}{k}}$

Also for the upper block

$f_{m a x} = μ m g = m \frac{k}{m + M} x$

$μ g = \frac{k}{m + M} (v_{0} \sqrt{\frac{m + M}{k}})$

$\Rightarrow \frac{k}{m + M} = {(\frac{μ g}{v_{0}})}^{2}$

$\Rightarrow k = {(\frac{μ g}{v_{0}})}^{2} (m + M)$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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