Q39E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 0.218 m/s

(b) 11.71N

Step 1: Identification of given data

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 ${\mu }_k=0.450$ 

Step 2: Significance of Newton’s 2 nd law of motion

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.

Step 3: (a) Determining the speed of each block after they move 3.00 cm

                           

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

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

Step 4: (b) Determining the tension in the cord

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 $11.71\mathrm{N}$