Question

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

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

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}{2 m} \ldots (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 $m a$ on the masses.
On applying the work-energy theorem, taking the centre of mass as a reference point.
$W_{\text {all forces }}=\Delta K$
$(F-m a) x_{1}+(m a) x_{2}-\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}=0$
$[\because$ w.r.t. centre of mass $\Delta K=0]$
$\Rightarrow \left(F-m \times \frac{F}{2 m}\right) x_{1}+\left(m \times \frac{F}{2 m}\right) x_{2}-\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}=0$
$\Rightarrow \left(F-\frac{F}{2}\right) x_{1}+\left(\frac{F}{2}\right) x_{2}-\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}=0$
$\Rightarrow \frac{F}{2} \times x_{1}+\frac{F}{2} \times x_{2}-\frac{1}{2} K\left(x_{1}+x_{2}\right)^{2}=0$
$\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, $F=10 \,N , K=2500\, N / m ^{2}$
So, $x_{\text {total }}=\frac{10}{2500} m=\frac{1}{250} m=\frac{100}{250} \,cm =0.4\, cm$
So, the maximum distance between the blocks will be $=$ natural length $+x_{\text {total }}$
$=10 \, cm +0.4 \, cm =10.4 \,cm $
[$ \because$ Natural length $=10 \, cm$ (given) ]