To find the minimum value of the tightening force F required to prevent belt slip, follow these steps:

1. Calculate the torque (τ) transmitted:

Power (P) is related to torque and angular velocity (ω) by the formula $P = \tau \cdot \omega$.
Given:
$P = 6.283\text{ kW} = 6283\text{ W}$
$N = 600\text{ RPM}$
$\omega = \frac{2 \pi N}{60} = \frac{2 \pi \cdot 600}{60} = 20 \pi\text{ rad/s} \approx 62.83\text{ rad/s}$

$\tau = \frac{P}{\omega} = \frac{6283}{62.83} = 100\text{ N}\cdot\text{m}$

2. Determine the difference between belt tensions ($T_1 - T_2$):

Torque is also given by the formula $\tau = (T_1 - T_2) \cdot r$.
Given radius $r = 20\text{ cm} = 0.2\text{ m}$:
$100 = (T_1 - T_2) \cdot 0.2$
$T_1 - T_2 = \frac{100}{0.2} = 500\text{ N}$ — (Eq. 1)

3. Use the tension ratio formula for belt slip:

At the point of slipping, the ratio of tensions is $\frac{T_1}{T_2} = e^{\mu \theta}$.
Given coefficient of friction $\mu = 0.3$ and angle of wrap $\theta = \pi$ radians (from the figure, the belt covers half the pulley):
$\frac{T_1}{T_2} = e^{0.3 \cdot \pi} \approx e^{0.9425} \approx 2.566$
$T_1 = 2.566 \cdot T_2$ — (Eq. 2)

4. Solve for $T_1$ and $T_2$:

Substitute Eq. 2 into Eq. 1:
$2.566 T_2 - T_2 = 500$
$1.566 T_2 = 500$
$T_2 = 319.28\text{ N}$
$T_1 = 319.28 + 500 = 819.28\text{ N}$

5. Calculate the tightening force F:

From the horizontal force equilibrium shown in the free-body diagram of the pulley:
$F = T_1 + T_2$
$F = 819.28 + 319.28 = 1138.56\text{ N}$

Convert to kN: $F = 1.13856\text{ kN}$

Rounding to 2 decimal places, the minimum value of the tightening force F is 1.14 kN.