Question

System shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of $5kg$ block when $2kg$ block leaves the contact with ground is (Force constant of spring $k=40N/m$ and $g=10m/s^2$ )

• A. $\sqrt{2}m/s$
• B. $2\sqrt{2}m/s$
• C. $2m/s$
• D. $4\sqrt{2}m/s$

Answer 1

The correct option is B $2\sqrt{2}m/s$

As soon as $2$ kg block leaves the ground then there will be same tension 'T' in the string such that

$T=20N$...(i)

and

$T=F_s$ ...(ii)

where,

$F_s=K{x}_0$...(iii)

where,

$x_o\to$ expansion in spring due to decrease in height of 5 kg block from the ground.

from (i), (ii) & (iii)

$\Rightarrow K{x}_o=20N$

$40x_o=20$

$\Rightarrow x_0=2m$

Apply energy conserration

Loss in (P.E. of block)= gain in (P.E. of spring +K.E of block)

$\Rightarrow mg{x}_0=\frac{1}{2}K{x}_0^2+\frac{1}{2}m{v}^2$

$\Rightarrow 5\times 10\times 2=\frac{1}{2}\times 40\times (2{)}^2+\frac{1}{2}\times 5\times v^2$

$100=80+\frac{5v^2}{2}$

$\Rightarrow \frac{5v^2}{2}=20$

$\Rightarrow v^2=\frac{40}{5}=8$

$\Rightarrow v=\sqrt{2}$

$\Rightarrow \Rightarrow v=2\sqrt{2}m{s}^{-1}$