Question

Two blocks $1$ and $2$ of equal mass $m$ are connected by an ideal string (see figure below) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants $k_1$ and $k_2$ such that $k_1>k_2$. Initially, both springs are unstretched . Block $1$ is slowly pulled down a distance $x$ and released. Just after the re lease the possible values of the magnitudes of the accelerations of the blocks $a_1$ and a $a_2$ can be

• A. either $\left(a_1=a_2=\frac{\left(k_1+k_2\right)x}{2m}\right)$ or $(a_1=\frac{k_{1}x}{m}-g$ and $a_2=\frac{k_{2}x}{m}+g)$
• B. $\left(a_1=a_2=\frac{\left(k_1+k_2\right)x}{2m}\right)$ only
• C. $\left(a_1=a_2=\frac{\left(k_1-k_2\right)x}{2m}\right)$ only
• D. either $\left(a_1=a_2=\frac{\left(k_1-k_2\right)x}{2m}\right)$ or $\left(a_1=a_2=\frac{\left(k_1{k}_2\right)x}{\left(k_1+k_2\right)m}-g\right)$

Answer 1

Answer: (B)

Free body diagram for block $1$ and $2$ are as shown below.

So, equations of force balance are
$T+k_{1}x-mg=m{a}_1$
and $k_{2}x+mg-T=m{a}_2$
As string is in extensible,
$a_1=a_2=a$ (let)
Then adding both equations, we have
$\left(k_1+k_2\right)x=2ma$
$\Rightarrow a=\left(\frac{k_1+k_2}{2m}\right).x$