Question

What will be the reading of spring balance shown in the figure, if it is moving in upward direction with a constant acceleration $a=2{\text{m/s}}^2$ and an initial velocity $v=2\text{m/s}$. (Take $g=10{\text{m/s}}^2$)

• A. $80\text{N}$
• B. $100\text{N}$
• C. $120\text{N}$
• D. $140\text{N}$

Answer 1

The correct option is C $120\text{N}$

The reading of spring balance will be equal to the spring force generated due to extension in spring. let the spring force be ($T$).

$\Rightarrow$From the frame of reference of lift, acceleration of block will be $\text{zero}$, and it will be in equilibrium.


Here $F_{\text{Pseudo}}=ma$, applied on block in a direction opposite to acceleration of lift.
$\Rightarrow$Applying equilibrium condition on block as per FBD:
$T=mg+ma$
$T=(10\times 10)+(10\times 2)$
$\therefore T=120\text{N}$
Hence, Spring balance reading will be $120N$