A block of mass M slides on a frictionless surface with an initial speed of v0 . 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

Question

A block of mass M slides on a frictionless surface with an initial speed of v0 . 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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-2s9qb5zwsafu.png" /> (A) ((μ g/v0))2((m + M)2/m) (B) ((μ g/v0))2(m + M) (C) ((μ g/v0))2((m + M)2/M) (D) ((μ g/v0))2(M - m)

Tardigrade Admin · Accepted Answer

(1/2)(M + m)(v0)2=(1/2)kx2 x=v0√(M + m/k) Also for the upper block fm a x=μ mg=m(k/m + M)x μ g=(k/m + M)(v0 √(m + M/k)) ⇒ (k/m + M)=((μ g/v0))2 ⇒ k=((μ g/v0))2(m + M)

$(\frac{μg}{v _{0}})^{2} \frac{( m + M ) ^{2}}{m}$

$(\frac{μg}{v _{0}})^{2} (m + M)$

$(\frac{μg}{v _{0}})^{2} \frac{( m + M ) ^{2}}{M}$

$(\frac{μg}{v _{0}})^{2} (M - m)$

(v0​μg​)2m(m+M)2​

(v0​μg​)2(m+M)

(v0​μg​)2M(m+M)2​

(v0​μg​)2(M−m)

21​(M+m)(v0​)2=21​kx2 x=v0​kM+m​​ Also for the upper block fmax​=μmg=mm+Mk​x μg=m+Mk​(v0​km+M​​) ⇒m+Mk​=(v0​μg​)2 ⇒k=(v0​μg​)2(m+M)

$(\frac{μg}{v _{0}})^{2} \frac{( m + M ) ^{2}}{m}$

$(\frac{μg}{v _{0}})^{2} (m + M)$

$(\frac{μg}{v _{0}})^{2} \frac{( m + M ) ^{2}}{M}$

$(\frac{μg}{v _{0}})^{2} (M - m)$