Figure shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood’s machine.One block has mass${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{kg}$; the other has mass${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kg}$. (a) What is the magnitude of the blocks’ acceleration and

(b) What is the tension in the cord?

1) The mass of the first block is,$m_1=1.30\mathrm{kg}$

2) The mass of the second block is,$m_2=2.80\mathrm{kg}$

The free body diagram gives us the idea of the direction of the forces acting on the body. The kinematic equations give the relationship between the initial velocity, final velocity, acceleration, displacement, and time.

Using the free body diagram of both blocks, we can write the equation of motion for each block.

By rearranging the two force equations we can get the equation for tension and acceleration.

$$\begin{gathered} T-m_{1}g=m_{1}a \\ {m}_{2}g-T=m_{2}a \\ a=\left(\frac{m_2-m_1}{m_1+m_2}\right)g \\ T=\left(\frac{2m_1{m}_2}{m_1+m_2}\right)g \end{gathered}$$

The free body diagrams of both blocks are as follows,

So, applying Newton’s second law we get the force equations,

$$\begin{gathered} T-m_{1}g=m_{1}a \\ {m}_{2}g-T=m_{2}a \end{gathered}$$

By rearranging these two equations, we get the equation for tension T and acceleration a, as

$$\begin{gathered} T=\left(\frac{2m_1{m}_2}{m_1+m_2}\right)g \\ a=\left(\frac{m_2-m_1}{m_1+m_2}\right)g \end{gathered}$$

By using the equation for acceleration, we get

$$\begin{gathered} \mathrm{a}=\left(\frac{2.8\mathrm{kg}-1.3\mathrm{kg}}{1.3\mathrm{kg}+2.8\mathrm{kg}}\right)9.8 \\ =3.59\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the block is $3.59\mathrm{m}/{\mathrm{s}}^2$.

To find the tension, let us use equation of tension.

$$\begin{gathered} T=\left(\frac{2{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right)\mathrm{g} \\ =\left(\frac{2\times 2.8\mathrm{kg}\times 1.3\mathrm{kg}}{2.8\mathrm{kg}+1.3\mathrm{kg}}\right)9.8\mathrm{m}/{\mathrm{s}}^2 \\ =17.4\mathrm{N} \end{gathered}$$

Therefore, the tension in the cord is 17.4 N.