91Ó°ÊÓ

Weight of block A is ${\mathrm{m}}_{\mathrm{A}}\mathrm{g}=45\mathrm{N}$

Weight of block B is${\mathrm{m}}_{\mathrm{B}}\mathrm{g}=25\mathrm{N}$

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

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

…(¾±)

For block A

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

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

From equation (i)

$$\begin{gathered} {\mathrm{μ}}_{k}N=m_{B}g \\ {\mathrm{μ}}_{k}N=\frac{m_{B}g}{N} \\ =\frac{m_{B}g}{m_{A}g} \end{gathered}$$

Substituting all the values in above equation

$$\begin{gathered} {\mathrm{μ}}_k=\frac{25N}{45N} \\ =0.56 \end{gathered}$$

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

A cat weight of 45.0 N , falls asleep on top of block A, so the total weight of block A ($m_{A}g$) will be 45 N + 45 N = 90 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)

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

Substitute all the values in above equation

$$\begin{gathered} \mathrm{a}=\frac{25\mathrm{N}-0.56×90\mathrm{N}}{{\displaystyle \frac{25\mathrm{N}}{\mathrm{g}}}+{\displaystyle \frac{90\mathrm{N}}{\mathrm{g}}}} \\ =\frac{25.4\mathrm{g}}{115} \\ =-2.16\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

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