91Ó°ÊÓ

Weight of block A is$m_{A}g=45 \text{N} $

Weight of block B is$m_{B}g=25 \text{N} $

Newton's second law states that an object will accelerate in the direction of the net force. This acceleration leads the object to begin moving more slowly before coming to a complete stop since the force of friction acts in the opposite direction from that of motion.

Block B is moving with constant speed so the acceleration will be and the net force also zero, so from free body diagram equate the horizontal and vertical forces

$$\begin{gathered} T-m_{B}g=∑F_{net} \\ T-m_{B}g=0 \end{gathered}$$

$T=m_{B}g$

For block A

Normal force on block A is$N=m_{A}g$

$$\begin{gathered} T-{\mathrm{μ}}_{K}N=∑F_{net} \\ T-{\mathrm{μ}}_{K}N=0 \\ T={\mathrm{μ}}_{K}N \end{gathered}$$

From equation (i)

$\begin{array}{rcl}{\mathrm{μ}}_{K}N& =& m_{B}g\\ {\mathrm{μ}}_K& =& \frac{m_{B}g}{N}\\ & =& \frac{m_{B}g}{m_{A}g}\end{array}$

Substituting all the values in above equation

$\begin{array}{rcl}{\mathrm{μ}}_K& =& \frac{25 \text{N}}{45 \text{N}}\\ & =& 0.56\end{array}$

So the coefficient of kinetic friction between block A and the tabletop is0.56.

A cat weight of , falls asleep on top of block A, so the total weight of block A ($m_{A}g$) will be

$45 \text{N}+45 \text{N}=90 \text{N}$

By applying Newton’s law on block A

$T-{\mathrm{μ}}_{K}N=m_{A}a$ …(¾±¾±)

Where T tension on rope is, a is acceleration and N is normal force

By applying Newton’s law on block B

$m_{B}g-T=m_{B}a$ …(¾±¾±¾±)

From equation (ii) and equation (iii)

$$\begin{gathered} m_{B}g-({\mathrm{μ}}_{K}N+m_{A}a)=m_{B}a \\ {m}_{B}a+m_{A}a=m_{B}g-{\mathrm{μ}}_{K}N \\ a=\frac{m_{B}g-{\mathrm{μ}}_K{m}_{A}g}{m_B+m_A} \end{gathered}$$

Substitute all the values in above equation

$\begin{array}{rcl}a& =& \frac{25 \text{N}-0.56×90 \text{N}}{\frac{25 \text{N}}{g}+\frac{90\text{N}}{g}}\\ & =& \frac{-25.4g}{115}\\ & =& -2.16 {\text{m/s}}^{\text{2}}\end{array}$

Hence the acceleration of block B is$-2.16m/s^2$ (-ve sign indicate that block B is moving upward direction).