Question

Sol.

kx₀ = mg

$x_{0} = \frac{mg}{k}$

compression from MP = Elongation from MP = x

From W.E.T. ⇒ mg(x + x₀) + $\frac{1}{2}k\left\lbrack O^{2} - (x + x_{0})^{2} \right\rbrack = 0 - \frac{1}{2}mv_{0}^{2} = - mgh$

⇒ $\frac{1}{2}k(x + x_{0})^{2} - mg(x + x_{0}) - {mgh} = 0 \Rightarrow (x + x_{0})$

From (1) kx – kx₀ = 2mg ⇒ kx – $k.\frac{mg}{k}$ = 9mg

⇒ x = $\frac{3mg}{k}$

∴ x + x₀ = $\frac{3mg}{k} + \frac{mg}{k} = \frac{4mg}{k}$

Now from W.E.T. ⇒ from NLP to max compression

mg (x + x₀) + $\frac{1}{2}k\left\lbrack O^{2} - (x + x_{0})^{2} \right\rbrack = 0 - \frac{1}{2}mv_{0}^{2}$

⇒ mg. $\frac{4mg}{k} - \frac{1}{2}k.\frac{(4{mg})^{2}}{k^{2}} = - \frac{1}{2}mv_{0}^{2}$

⇒ $- \frac{4({mg})^{2}}{k} = - \frac{1}{2}mv_{0}^{2}$

⇒ $\frac{8m^{2}g^{2}}{k} = m.2gh \Rightarrow h = \frac{4mg}{k}$

Alternate Solution

$\frac{1}{2}m(2{gh}) = {mgh} + \frac{1}{2}kx^{2}$ ...(i)

kx = 2 mg ...(ii)

From (i) & (ii)

$h = \frac{4mg}{k}$