Question

A block with mass $M=6.0\mathrm{k}\mathrm{g}$ rests on a frictionless table and is attached by a horizontal spring $\left(k=130\mathrm{N}/\mathrm{m}\right)$ to a wall. A second block, of mass $m=1.25\mathrm{k}\mathrm{g}$,rests on top of $M$. The coefficient of static friction between the two blocks is $0.30$. What is the maximum possible amplitude of oscillation such that $m$ will not slip off $M$?

Short answer

The maximum possible amplitude of oscillation is 0.16 m.

Step-by-step solution

Concept of acceleration applied on frictional surface

Acceleration applied on the frictional surface is calculated using the coefficient of static friction and normal reaction, which ultimately equivalent to the weight of the object.

Given data

The mass of the block is$M=6.0\mathrm{k}\mathrm{g}$.

The mass of the second block is$m=1.25\mathrm{k}\mathrm{g}$.

The spring stiffness constant is$k=130\mathrm{N}/\mathrm{m}$.

The coefficient of static friction is${\mu }_{\mathrm{s}}=0.30$.

Calculation of maximum amplitude

In the stationary situation, static friction will apply in between both of the blocks. This frictional force causes block to move upon the bigger block M. Thus, equating both of the forces due to acceleration and static friction as:

$\begin{array}{r}cF=F_{\mathrm{f}}\\ m{a}_{\mathrm{m}\mathrm{a}\mathrm{x}}={\mu }_{\mathrm{s}}mg\\ a_{\mathrm{m}\mathrm{a}\mathrm{x}}={\mu }_{\mathrm{s}}g\end{array}$

Here,$F_{\mathrm{f}}$is the frictional force, F is the force due to acceleration,$a_{\mathrm{m}\mathrm{a}\mathrm{x}}$is the maximum acceleration and$g$is the acceleration due to gravity.

Now, for this second block to stay stationary on the bigger block without slipping, maximum acceleration of bigger block of mass$M$is also$a_{\mathrm{m}\mathrm{a}\mathrm{x}}={\mu }_{\mathrm{s}}g$. However, due to connection with spring, system of block will perform simple harmonic motion, then the maximum acceleration is calculated as:

$\begin{array}{r}c{a}_{\mathrm{m}\mathrm{a}\mathrm{x}}={\omega }^{2}A\\ =\frac{k}{m+M}A\end{array}$

Here,$\omega$is the angular frequency and A is the amplitude.

Equating this equation with the previous expression of maximum acceleration as:

$\begin{array}{rl}c{\mu }_{\mathrm{s}}g& =\frac{k}{M}A\\ A& =\frac{{\mu }_{\mathrm{s}}g}{k}\left(m+M\right)\\ A& =\frac{\left(0.30\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(130\mathrm{N}/\mathrm{m}\right)}\left(6.0\mathrm{k}\mathrm{g}+1.25\mathrm{k}\mathrm{g}\right)\\ A& =0.16\mathrm{m}\end{array}$

Hence, the maximum amplitude of the oscillations is 0.16 m.