For Question , calculat the block's speed asit passes through the equilibrium point ?
Hint `: E = K+U = (1)/(2) mv^(2) + (1)/(2) kx^(2) = (1)/(2) mv^(2) +0` . Calculate v

Explanation

We want the speed at `x=0` where the PE is `U = (1)/(2) kx^(2) = 0` and the mechanical energy is entirely kinetic energy ,
So, `E = K+U= (1)/(2)mv^(2) +(1)/(2)kx^(2)+= (1)/(2) mv^(2) +0`
`8.588 xx 10^(7) = (1)/(2) xx ( 5.44 xx 10^(5) kg )v^(2)`
` v = 17.77 m//s`