Suppose that in Fig. 5-12, the masses of the blocks are 2.0 kg and 4.0 kg. (a) Which mass should the hanging block have if the magnitude of the acceleration is to be as large as possible? What then are (b) the magnitude of the acceleration and (c) the tension in the cord?

Fig. 5-12

$$\begin{gathered} m_1=2kg \\ {m}_2=4kg \end{gathered}$$

Using the concept of Newton’s second law of motion, we can decide which mass should be hanging. We can also write two equations for two masses in terms of tension and acceleration, i.e. two unknown variables. Tension and acceleration are common for both the objects. So these two equations can be solved for the acceleration and tension in the string.

Formula

${\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathit{a}$

Here,${\mathbf{F}}_{\mathbf{net}}$ is the force, m is the mass of the object, and a is the acceleration of the object.

(a)

Heavier mass should be hanging for maximum acceleration as net force is greater for heavier mass, so 4 mass should be hanging.

(b)

Net force on mass $m_1$is as follow

$F_{net}=T$

But according to newton’s second law,

$F_{net}=m_1\times a$

So

role="math" localid="1660911650209" $T=m_1\times a$

Now net force on second mass$m_2=2\mathrm{kg}$

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

Rearrange this equation to find acceleration.

$a=\frac{m_{2}g}{m_1+m_2}$

Substitute the values in the above equation to calculate acceleration

$\begin{array}{l}a=\frac{4\mathrm{kg}\times 9.811\mathrm{m}/{\mathrm{s}}^2}{2\mathrm{kg}+4\mathrm{kg}}\\ =6.54\mathrm{m}/{\mathrm{s}}^2\end{array}$

(c)

Tension in chord is calculated using Newton’s law as,

$\begin{array}{l}T=m_1\times a\\ =2\times 6.54m/s^2\\ =13.08N\end{array}$

Therefore, the tension in the cord is 13 N.