$A$ block of mass $1 \; kg$ is fastened to a spring with a spring constant of $50 \; N m^{-1}$. The block is pulled to a distance $x = 10 \; cm$ from its equilibrium position at $x = 0$ on a frictionless surface and released from rest at $t = 0$. Calculate the kinetic,potential,and total energies of the block when it is $5 \; cm$ away from the mean position.

(N/A) The block executes $SHM$. The angular frequency is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{1}} = 7.07 \; rad \; s^{-1}$.
Given amplitude $A = 0.1 \; m$ and displacement $x = 0.05 \; m$.
The potential energy $(P.E.)$ at $x = 0.05 \; m$ is:
$P.E. = \frac{1}{2} k x^2 = \frac{1}{2} \times 50 \times (0.05)^2 = 25 \times 0.0025 = 0.0625 \; J$.
The total energy $(E)$ of the system is constant and equal to the potential energy at maximum displacement $(x = A)$:
$E = \frac{1}{2} k A^2 = \frac{1}{2} \times 50 \times (0.1)^2 = 25 \times 0.01 = 0.25 \; J$.
The kinetic energy $(K.E.)$ at $x = 0.05 \; m$ is:
$K.E. = E - P.E. = 0.25 - 0.0625 = 0.1875 \; J \approx 0.19 \; J$.