Question

The coefficient of kinetic friction between the $2.0kg$block in FIGURE $P7.38$ and the table is $0.30.$ What is the acceleration of the $2.0kg$ block?

Short answer

Acceleration of the block is$6.91m/s^2$.

Step-by-step solution

Given information

Coefficient of friction between $2kg$block and table, $\mu =0.3$

Given the figure of three blocks

Explanation

Consider the net force applied by $1kg$block

localid="1649647806258" $$\begin{gathered} F_1=\left(1kg\right)\times \left(9.81m/s^2\right) \\ =9.81N\left(\text{righttoleft}\right) \end{gathered}$$

The friction force applied to the block $2$

localid="1649647844745" $$\begin{gathered} F_2=\left(2kg\right)\times 0.3\times \left(9.81m/s^2\right) \\ =5.8N\left(\text{righttoleft}\right) \end{gathered}$$

The force applied by the block $3$

localid="1649647866757" $$\begin{gathered} F_3=\left(3kg\right)\times \left(9.81m/s^2\right) \\ =29.43N\left(\text{lefttoright}\right) \end{gathered}$$

Net force o block $2$

localid="1649647898415" $$\begin{gathered} F=\left(29.43N\right)-\left(5.8N\right)-\left(9.81N\right) \\ =13.82N \end{gathered}$$

The acceleration of the block is calculated as

localid="1649647916525" $$\begin{gathered} F=ma \\ 13.82N=2kg\times a \\ a=6.91m/s^2 \end{gathered}$$

Acceleration of the block is$6.91m/s^2$.