The bottom crate is pulled by a force of 20 N to the right. The μk between the bottom crate and the surface is 0.32. The μk between the bottom crate and the top crate is also 0.32. (Figure 1) (below) Find the acceleration of the 2.0 kg crate.

Given data:

• The force acting on the bottom crate is F=20 N.
• The coefficient of friction between bottom crate and the surface is μ_k=0.32.
• The coefficient of friction between bottom crate and top crate is μ_k=0.32.
• The mass of the upper crate is m₁=1 kg.
• The mass of the bottom crate is m₂=2 kg.

The representation of various forces acting on upper crate is shown as,

The expression for the tension in the string due to upper crate can be calculated as,

$$\begin{gathered} T={\mu }_{\mathrm{k}}{N}_1+m_{1}a \\ T=\left({\mu }_{\mathrm{k}}{m}_{1}g\right)+m_{1}a \end{gathered}$$

Here, N₁ is the normal force of upper crate, and g is the gravitational acceleration having standard value of 9.81 m/s².

Substitute the known values in the above expression.

$$\begin{gathered} T=\left(0.32\times 1 \mathrm{kg}\times 9.81 \mathrm{m}/{\mathrm{s}}^2\right)+\left(1 \mathrm{kg}\times a\right) \\ T=3.14 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2+\left(1 \mathrm{kg}\times a\right) .........\left(1\right) \end{gathered}$$

The representation of various forces acting on bottom crate is shown as,