Question

A block of mass $m\text{kg}$ is attached to a massless spring of spring constant $k\text{N/m}$. This system is accelerated upwards with an acceleration $a{\text{m/s}}^2$. Then the elongation in spring will be (in metres):

• A. $\frac{mg}{k}$
• B. $\frac{m(g-a)}{k}$
• C. $\frac{m(g+a)}{k}$
• D. $\frac{ma}{k}$

Answer 1

The correct option is C $\frac{m(g+a)}{k}$


From the FBD of block, applying Newton's $2^{nd}$ law along direction of acceleration:
$T-mg=ma$......(1)
$T=mg+ma$ ......(2)

Let the elongation in spring be $x\text{m}$, hence spring force $kx$ will act in the form of tension $T$ on the block.
$\Rightarrow kx=T$ ........(3)

From Eq.(2) and Eq.(3), we get:

$kx=m(g+a)$
$\therefore x=\frac{m(g+a)}{k}$