Question

A block of mass $m=1kg$ placed on top of another block of mass $M=5kg$ is attached to a horizontal spring of force constant $k=20N/m$ as shown in figure. The coefficient of friction between the blocks is $\mu$ where as the lower block slides on a frictionless surface. The amplitude oscillation is 0.4 m. What is the minimum value of $\mu$ such that the upper block does not slip over the lower block?

• A. 0.133
• B. 0.5
• C. 0.362
• D. 0.21

Answer 1

Answer: (A)

0.133

The upper block does not slip over the lower block when the restoring force is balanced by the friction force of lower block against ground. i.e, $k{x}_0=\mu (M+m)g$
or $\mu =\frac{k{x}_0}{(M+m)g}=\frac{20\times 0.4}{(5+1)10}=0.133$