The sketch below shows a tow-truck of mass 2500 kg pulling a car of mass 800 kg up a rough inclined surface which makes an angle of 30° with the horizontal.

The normal force on an inclined plane is the component of the gravitational force perpendicular to the surface.

$$F_N = m_{car} g \cos\theta$$ $$F_N = (800 \text{ kg})(9.8 \text{ m/s}^2) \cos(30^\circ)$$ $$F_N = 7840 \text{ N} \times 0.8660$$ $$F_N = 6790.99 \text{ N}$$

The kinetic frictional force is given by the product of the coefficient of kinetic friction and the normal force.

$$F_f = \mu_k F_N$$ $$F_f = (0.25)(6790.99 \text{ N})$$ $$F_f = 1697.75 \text{ N}$$

The kinetic frictional force is $\boxed{\text{1698 N}}$.

4.1.4 Calculate the tension T in the cable.

Given:
Mass of car ($m_{car}$) = $800 \text{ kg}$
Mass of truck ($m_{truck}$) = $2500 \text{ kg}$
Net force on truck ($F_{net, truck}$) = $1520 \text{ N}$
Angle of incline ($\theta$) = $30^\circ$
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$
Kinetic frictional force ($F_f$) = $1697.75 \text{ N}$ (from 4.1.3)

Since the truck is pulling the car with a cable, they both experience the same acceleration.

$$a = \frac{F_{net, truck}}{m_{truck}}$$ $$a = \frac{1520 \text{ N}}{2500 \text{ kg}}$$ $$a = 0.608 \text{ m/s}^2$$

The net force on the car along the incline is the sum of the forces acting parallel to the incline.

$$F_{net, car} = T - m_{car} g \sin\theta - F_f$$

According to Newton's Second Law, $F_{net, car} = m_{car} a$.

$$m_{car} a = T - m_{car} g \sin\theta - F_f$$

$$T = m_{car} a + m_{car} g \sin\theta + F_f$$ $$T = (800 \text{ kg})(0.608 \text{ m/s}^2) + (800 \text{ kg})(9.8 \text{ m/s}^2) \sin(30^\circ) + 1697.75 \text{ N}$$ $$T = 486.4 \text{ N} + (7840 \text{ N})(0.5) + 1697.75 \text{ N}$$ $$T = 486.4 \text{ N} + 3920 \text{ N} + 1697.75 \text{ N}$$ $$T = 6104.15 \text{ N}$$

The tension in the cable is $\boxed{\text{6104 N}}$.

4.2.1 State Newton's Law of universal gravitation in words.

Newton's Law of Universal Gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

4.2.2 Calculate the distance between the spaceship and the centre of Earth when Earth exerts a force of 2458 N on the spaceship.

Given:
Force ($F$) = $2458 \text{ N}$
Mass of spaceship ($m_{spaceship}$) = $1250 \text{ kg}$
Gravitational constant ($G$) = $6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$
Mass of Earth ($M_E$) = $5.972 \times 10^{24} \text{ kg}$

$$r^2 = G \frac{m_{spaceship} M_E}{F}$$ $$r = \sqrt{G \frac{m_{spaceship} M_E}{F}}$$

$$r = \sqrt{\frac{(6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2)(1250 \text{ kg})(5.972 \times 10^{24} \text{ kg})}{2458 \text{ N}}}$$ $$r = \sqrt{\frac{4.978355 \times 10^{17} \text{ N}\cdot\text{m}^2}{2458 \text{ N}}}$$ $$r = \sqrt{2.02536 \times 10^{14} \text{ m}^2}$$ $$r = 1.42315 \times 10^7 \text{ m}$$

The distance between the spaceship and the center of Earth is \boxed{\text{1.42 \times 10^7 m}}.