Question

Find the reading (in $\text{newtons}$) of spring balance when system shown in the figure below is released from rest. $($Take $g=10\text{m/}{s}^2)$

• A. $300\text{N}$
• B. $\frac{200}{3}\text{N}$
• C. $\frac{400}{3}\text{N}$
• D. $\frac{800}{3}\text{N}$

Answer 1

The correct option is D $\frac{800}{3}\text{N}$

Considering the $40\text{kg}$ block, $20\text{kg}$ block and spring as one system,

$a_{system}=\frac{\text{Net pulling force}}{\text{Total mass}}$

$\Rightarrow a=\frac{40g-20g}{40+20}$

$\therefore a=\frac{20\times 10}{60}=\frac{10}{3}{\text{m/s}}^2$

The forces on both ends of spring will be same.


This spring force $T$ will be the reading of spring balance.

From $\text{F.B.D}$ of $40\text{kg}$:


From Newtons second law,

$40g-T=40a$

$\Rightarrow 40\times 10-T=40\times \frac{10}{3}$

$\therefore T=\frac{800}{3}\text{N}$

Hence, option (d) is the correct answer.