As shown in the figure, a spring with constant k = 100 N/m is...

At the equilibrium position, the spring force balances the constant external force $F$. Let the compression in the spring be $x_0$.

$k x_0 = F \Rightarrow 100 x_0 = 10 \Rightarrow x_0 = 0.1 \text{ m} = 10 \text{ cm}$

The angular frequency of the SHM is $\omega = \sqrt{\dfrac{k}{M}} = \sqrt{\dfrac{100}{1}} = 10 \text{ rad/s}$.

The time period is $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} = \dfrac{\pi}{5} \text{ s}$.

Since the blow is given at the equilibrium position, the imparted speed is the maximum velocity of the SHM.

v_\max = A\omega \Rightarrow 2 = A \times 10 \Rightarrow A = 0.2 \text{ m} = 20 \text{ cm}

The block oscillates about the equilibrium position ($10 \text{ cm}$ compression) with an amplitude of $20 \text{ cm}$.

At the left extreme, the spring is further compressed. The maximum compression is $x_{left} = x_0 + A = 10 \text{ cm} + 20 \text{ cm} = 30 \text{ cm} = 0.3 \text{ m}$.

The potential energy of the spring at the left extreme is $U_{left} = \dfrac{1}{2} k x_{left}^2 = \dfrac{1}{2} \times 100 \times (0.3)^2 = 4.5 \text{ J}$.

At the right extreme, the position of the block is $x_{right} = x_0 - A = 10 \text{ cm} - 20 \text{ cm} = -10 \text{ cm}$. This corresponds to an extension of $10 \text{ cm} = 0.1 \text{ m}$.

The potential energy of the spring at the right extreme is $U_{right} = \dfrac{1}{2} k x_{right}^2 = \dfrac{1}{2} \times 100 \times (-0.1)^2 = 0.5 \text{ J}$.

Answer: The time period of the resulting simple harmonic motion is $\dfrac{\pi}{5} \text{ s}$.; The amplitude of the resulting simple harmonic motion is $20 \text{ cm}$.; The potential energy of the spring when the block is at the left extreme is $4.5 \text{ J}$.