Question

In the figure shown, block $A$ is displaced by $0.75\text{m}$ to the right and block $B$ is displaced by $0.25\text{m}$ to left. Find the reading of the spring balance. Take $g=10m/s^2$

• A. $10\text{kg}$
• B. $20\text{kg}$
• C. $100\text{kg}$
• D. $200\text{kg}$

Answer 1

The correct option is B $20\text{kg}$

Let the displacement of the two blocks are as shown.
Here spring constant $k=200N/m$

As we know that the spring balance reads the net spring force applied by the spring on the body in order to get restored.
So, from the figure shown above we can say that
Spring force on body $A$ is $F_{sA}=k{x}_1=200\times 0.75=150N$
Similarly, spring force on body $B$ is $F_{sB}=k{x}_2=200\times 0.25=50N$
So, net force applied by the spring is $F_{net}=150+50=200N$
Thus, the equivalent mass measured by the spring balance is $\frac{F_{net}}{g}=\frac{200}{10}=20Kg$

Alternate:

Net displacement of the spring, $\delta x=(0.75+0.25)=1\text{m}$
Spring force, $F_s=k.\delta x=200\times 1=200\text{N}$
$\therefore \text{Equivalent mass}=\frac{F}{g}=\frac{200}{10}=20\text{kg}$