Let $x_f$ be the final position where the block stops. The block starts and ends at rest, so the total change in kinetic energy is zero, $\Delta K = 0$.

By the work-energy theorem, the net work done on the block is equal to the change in its kinetic energy:

$$W_{net} = W_F + W_s = \Delta K = 0$$

The work done by the constant applied force $F$ is $W_F = Fx_f$. The work done by the spring is $W_s = -\frac{1}{2}kx_f^2$.

[Q1] Substitute the work expressions into the work-energy theorem:

$$Fx_f - \frac{1}{2}kx_f^2 = 0$$

Since $x_f eq 0$, we can divide by $x_f$ to find the final position:

$$F - \frac{1}{2}kx_f = 0 \implies x_f = \frac{2F}{k}$$

[Q2] The work done by the applied force is:

$$W_F = Fx_f = F\left(\frac{2F}{k}\right) = \frac{2F^2}{k}$$

[Q3] The work done by the spring force is:

$$W_s = -\frac{1}{2}kx_f^2 = -\frac{1}{2}k\left(\frac{2F}{k}\right)^2 = -\frac{2F^2}{k}$$

Note that $W_F + W_s = 0$, as expected.

[Q4] The kinetic energy is maximum when the net force on the block is zero, as this is the point of maximum velocity.

$$F_{net} = F - F_s = F - kx = 0$$

Solving for the position $x_{maxK}$ where kinetic energy is maximum:

$$x_{maxK} = \frac{F}{k}$$

[Q5] To find the maximum kinetic energy $K_{max}$, we apply the work-energy theorem from the start ($x=0$) to the position $x_{maxK}$. The initial kinetic energy is $K_i=0$.

$$K_{max} = W_{net}(0 \to x_{maxK}) = W_F + W_s$$ $$W_F = Fx_{maxK} = F\left(\frac{F}{k}\right) = \frac{F^2}{k}$$ $$W_s = -\frac{1}{2}kx_{maxK}^2 = -\frac{1}{2}k\left(\frac{F}{k}\right)^2 = -\frac{F^2}{2k}$$

The maximum kinetic energy is the sum of these work terms:

$$K_{max} = \frac{F^2}{k} - \frac{F^2}{2k} = \frac{F^2}{2k}$$