Physics - W.E.P. Question with Solution | TestHub

In equilibrium$\left({\text{F}}_{\text{net}}=0\right)$

$\text{T}=kx$........ (1)

$2\text{T}+\frac{\text{kx}}{2}=\text{mg}$........ (2)

Solving these equations, we have

$\text{x}=\frac{2\text{mg}}{5\text{k}}=\frac{2\times 1\times 10}{5\times 10}=\text{0.4 m}$



Let v be the speed of block placed horizontally, then

$\frac{1}{2}{\text{mv}}^2+\frac{1}{2}\text{m}{\left(\frac{\text{v}}{2}\right)}^2+\frac{1}{2}{\text{kx}}^2+\frac{1}{2}\text{k}{\left(\frac{\text{x}}{2}\right)}^2=mg\left(\frac{\text{x}}{2}\right)$

or$\frac{5}{8}{\text{mv}}^2=\frac{\text{mgx}}{2}-\frac{5}{8}{\text{kx}}^2$

or$\text{v}=\sqrt{\frac{8}{5}\left(\frac{\text{gx}}{2}-\frac{5{\text{kx}}^2}{8\text{m}}\right)}$

Substituting the values, we have

$\text{v}=\sqrt{\frac{8}{5}\left[\frac{10\times \text{0.4}}{2}-\frac{5\times 10\times \text{0.16}}{8\times 1}\right]}$

$=1.26\; m/s$