Question

Two blocks of equal mass are connected with a massless spring of spring constant $2500N/m^2$ and length $10cm$ at rest on the frictionless horizontal plane. If a constant horizontal force $10N$ is applied as shown in the figure, find the maximum distance between the blocks.

• A. $10.8cm$
• B. $10.4cm$
• C. $10.6cm$
• D. $10.0cm$

Answer 1

Answer: (B)

In the case of maximum elongation, whole system will be moving with a common acceleration $a$, so for the system

$F=(m+m)a$
$\Rightarrow a=\frac{F}{2m}\dots (i)$
The centre of mass of the system will also move with the same acceleration $a$.
Let us suppose that both the masses elongate the spring by a distance $x_1$ and $x_2$, respectively.
So, on taking the frame of centre of mass there will act a pseudo force equal to $ma$ on the masses.
On applying the work-energy theorem, taking the centre of mass as a reference point.
$W_{\text{all forces }}=\Delta K$
$(F-ma)x_1+(ma)x_2-\frac{1}{2}K{\left(x_1+x_2\right)}^2=0$
$[\because$ w.r.t. centre of mass $\Delta K=0]$
$\Rightarrow \left(F-m\times \frac{F}{2m}\right)x_1+\left(m\times \frac{F}{2m}\right)x_2-\frac{1}{2}K{\left(x_1+x_2\right)}^2=0$
$\Rightarrow \left(F-\frac{F}{2}\right)x_1+\left(\frac{F}{2}\right)x_2-\frac{1}{2}K{\left(x_1+x_2\right)}^2=0$
$\Rightarrow \frac{F}{2}\times x_1+\frac{F}{2}\times x_2-\frac{1}{2}K{\left(x_1+x_2\right)}^2=0$
$\Rightarrow \frac{F}{2}\left(x_1+x_2\right)=\frac{1}{2}K{\left(x_1+x_2\right)}^2$
$\Rightarrow F=K\left(x_1+x_2\right)$
$\Rightarrow x_1+x_2=\frac{F}{K}$
$\Rightarrow x_{\text{total }}=\frac{F}{K}$
Here, $F=10N,K=2500N/m^2$
So, $x_{\text{total }}=\frac{10}{2500}m=\frac{1}{250}m=\frac{100}{250}cm=0.4cm$
So, the maximum distance between the blocks will be $=$ natural length $+x_{\text{total }}$
$=10cm+0.4cm=10.4cm$
[$\because$ Natural length $=10cm$ (given) ]