Question

a mass less spring with a force constant K=40 n/m hangs vertically from the ceiling. A 0.2kg block is attached to the free end of the string and held in such a position that the spring has its natural length and suddenly released. The max elastic energy stored in the spring

Answer 1

answer in the above pic

Answer 2

Given that,

Force constant = 40 N/m

Mass = 0.2 kg

We need to calculate the length of the spring

Using conservation of mechanical energy

Here, change in mechanical energy is zero

So, $K.E+P.E=0$

$0+(P.E)_{g}+(P.E)_{x}=0$

$-mgx+\dfrac{1}{2}kx^2$

$\dfrac{1}{2}kx^2=mgx$

$x=\dfrac{2mg}{k}$

Put the value into the formula

$x=\dfrac{2\times0.2\times9.8}{40}$

$x=0.098\ m$

We need to calculate the max elastic energy stored in the spring

Using formula of energy

$E=\dfrac{1}{2}kx^2$

Put the value into the formula

$E=\dfrac{1}{2}\times40\times(0.098)^2$

$E=0.2\ J$

Hence, The max elastic energy stored in the spring is 0.2 J