As shown in Fig. E5.34, block A (mass 2.25 kg) rests on a tabletop. It is connected by a horizontal cord passing over a light, frictionless pulley to a hanging block B (mass 1.30kg ). The coefficient of kinetic friction between block A and the tabletop is 0.450. The blocks are released then from rest. Draw one or more free-body diagrams to find (a) the speed of each block after they move 3.00 cm and (b) the tension in the cord

Mass of block A is $m_A=2.25\mathrm{kg}$

Mass of block B is$m_B=1.30\mathrm{kg}$

Coefficient of kinetic friction between block A and the tabletop is ${\mu }_k=0.450$

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.

Expression of normal reaction force acting on block A is

$N=m_{A}g$

Expression of friction force acting on block A is

$$\begin{gathered} f={\mu }_{K}N \\ ={\mu }_K{m}_{A}g \end{gathered}$$

If both block released from rest then applying Newton’s force equation on block A

$$\begin{gathered} T-f=\underset{}{\sum F_{net}} \\ T-{\mu }_K{m}_{A}g=m_{A}a \\ T=m_{A}a+{\mu }_K{m}_{A}g \end{gathered}$$ …(i)

Applying Newton’s force equation on block B

$$\begin{gathered} m_{B}g-T=\underset{}{\sum F_{net}} \\ {m}_{B}g-T=m_{B}a \\ T=m_{B}g-m_{B}a \end{gathered}$$ …(ii)

From equation (i) and (ii)

$$\begin{gathered} m_{A}a+{\mu }_K{m}_{A}g=m_{B}g-m_{B}a \\ {m}_{A}a+m_{B}a=m_{B}g-{\mu }_K{m}_{A}g \\ a=\frac{\left(m_B-{\mu }_K{m}_A\right)g}{\left(m_A+m_B\right)} \end{gathered}$$

Substitute the values in above equation

$$\begin{gathered} a=\frac{\left(1.30\mathrm{kg}-0.45\times 2.25\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2}{\left(2.25\mathrm{kg}+1.30\mathrm{kg}\right)} \\ =0.793\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Distance moved by each block is $d=3\mathrm{cm}\mathrm{or}0.03\mathrm{m}$

So the speed of each block after they move 3cm is

$$\begin{gathered} v=\sqrt{2ad} \\ =\sqrt{2\times 0.793\mathrm{m}/{\mathrm{s}}^2\times 0.03\mathrm{m}} \\ =0.218\mathrm{m}/\mathrm{s} \end{gathered}$$

From equation (ii)

$$\begin{gathered} T=m_{B}g-m_{B}a \\ =\left(1.3\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)-\left(1.3\mathrm{kg}\times 0.793\mathrm{m}/{\mathrm{s}}^2\right) \\ =11.71\mathrm{N} \end{gathered}$$

So the tension in the cord is$11.71\mathrm{N}$