When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. Block 1 has mass M, block 2 has $\mathbf{2}\mathbf{M}$, and block 3 has $\mathbf{2}\mathbf{M}$. What is the coefficient of kinetic friction between block 2 and the table?

Acceleration,$\mathrm{a}=0.500\mathrm{m}/{\mathrm{s}}^2$

Mass of block 1,${\mathrm{M}}_1=\mathrm{M}$

Mass of block 2,${\mathrm{M}}_2=2\mathrm{M}$

Mass of block 3,${\mathrm{M}}_3=2\mathrm{M}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. First, draw the free body diagram for each box and then by using Newton's 2nd law of motion, coefficient of friction between the table and box can be found.

Formula:

$F_{net}=\sum Ma$

where, F is the force, Mis mass and a is an acceleration.

Free Body Diagram of the system:

By using Newton’s 2nd law of motion along the vertical direction for middle box having mass 2M,

$$\begin{gathered} N+2Mg=0 \\ N=2Mg \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_s={\mu }_{s}N \\ ={\mu }_s\left(2Mg\right) \end{gathered}$$

Now, applying Newton’s 2nd along horizontal and vertical direction for all 3 boxes,

$$\begin{gathered} T_{12}-Mg=Ma\left(i\right) \\ {T}_{23}-T_{12}-\left(2Mg{\mu }_s\right)=2Ma\left(ii\right) \\ 2Mg-T_{23}=2Ma\left(iii\right) \end{gathered}$$

By adding equations (i), (ii), and ,(iii)

Therefore, the coefficient of kinetic friction between block 2 and the table is 0.37.