Step 1: Analyzing Phase 2 (Retardation)

During the second phase (after force $F$ is removed), the only horizontal force acting on the block is kinetic friction $f$. The block travels a distance $d_2 = 1 \text{ m}$ before stopping.

Let the maximum velocity at the end of Phase 1 be $v_{max}$. Using the work-energy theorem for Phase 2:

$$W_{net} = \Delta K$$ $$-f \cdot d_2 = 0 – \frac{1}{2}mv_{max}^2 \quad \dots(1)$$

Alternatively, using kinematics ($v^2 – u^2 = 2as$):

$$0^2 – v_{max}^2 = 2(-a_2)d_2$$ $$v_{max}^2 = 2 a_2 (1) \quad \dots(2)$$

Step 2: Analyzing Phase 1 (Acceleration)

The block starts from rest and accelerates for $t_1 = 5 \text{ s}$, covering a distance $d_1 = 5 \text{ m}$.

$$d_1 = u t_1 + \frac{1}{2}a_1 t_1^2$$ $$5 = 0 + \frac{1}{2}a_1 (5)^2 \implies 5 = 12.5 a_1 \implies a_1 = 0.4 \text{ m/s}^2$$

Now we find the maximum velocity $v_{max}$ reached at the end of 5 seconds:

$$v_{max} = u + a_1 t_1 = 0 + 0.4(5) = 2 \text{ m/s}$$

This confirms statement (b) is correct.

Step 3: Calculating Friction and Force

Using $v_{max} = 2 \text{ m/s}$ in equation (2) for Phase 2:

$$2^2 = 2 a_2 (1) \implies 4 = 2 a_2 \implies a_2 = 2 \text{ m/s}^2$$

The deceleration is caused by friction $f = \mu mg$. Thus, $a_2 = \mu g$.

$$\mu g = 2 \implies \mu(10) = 2 \implies \mu = 0.2$$

This confirms statement (c) is correct.

Now, applying Newton’s Second Law for Phase 1:

$$F_{net} = ma_1$$ $$F – f = m a_1$$ $$F – \mu mg = m a_1$$ $$F – 0.2(5)(10) = 5(0.4)$$ $$F – 10 = 2 \implies F = 12 \text{ N}$$

This confirms statement (a) is correct.

Step 4: Time of Motion in Phase 2

The time $t_2$ taken to stop in Phase 2 can be found using $v = u + at$:

$$0 = v_{max} – a_2 t_2$$ $$0 = 2 – 2 t_2 \implies t_2 = 1 \text{ s}$$

This confirms statement (d) is correct.