Question

In the figure a smooth pulley of negligible weight is suspended by spring balance $1kg$ and $5kg$ are attached to the opposite ends of spring passing over pulley and move with acceleration because of gravity. During the motion, spring balance reads weight of ?

Answer 1

From the FBD of blocks we can write Newton's 2nd law equation as:

For 5 kg block: $5g-T=5a$

For 1 kg block : $T-g=a$

adding both equation we get value of $a=\frac{4g}{6}=\frac{2g}{3}\frac{m}{s^2}$

using this in any one of the above equation we get value of as $T=g+a=\frac{5g}{3}N$

Now, from FBD we can see that reading of spring balance will be $2T$, hence spring balance will read weight of $\frac{10g}{3}=33.34N$