(a) The work done against friction is calculated using the formula:

\(WD = F \times d\)

where \(F = 40 \text{ N}\) and \(d = 15 \text{ m}\). Thus,

\(WD = 40 \times 15 = 600 \text{ J}\)

(b) The change in gravitational potential energy is given by:

\(\Delta PE = mgh\)

where \(m = 5 \text{ kg}\), \(g = 10 \text{ m/s}^2\), and \(h = 15 \sin 20^{\circ}\). Therefore,

\(\Delta PE = 5 \times 10 \times 15 \times \sin 20^{\circ} \approx 257 \text{ J}\)

(c) The work done by the pulling force includes both the work against friction and the change in potential energy:

\(WD = 40 \times 15 + 5 \times 10 \times 15 \times \sin 20^{\circ} = 857 \text{ J}\)

(a) For particle A: Using Newton's second law, the tension in the string is given by:

\(T = 0.3a\)

For particle B: The forces acting on B are the tension, the component of gravitational force down the plane, and the applied force. Using Newton's second law:

\(3.5 + 0.5g \sin 30^{\circ} - T = 0.5a\)

For the system: Adding the equations for A and B:

\(3.5 + 0.5g \sin 30^{\circ} = (0.3 + 0.5)a\)

Solving for acceleration \(a\):

\(a = 7.5 \text{ m/s}^2\)

Substituting \(a\) back to find tension \(T\):

\(T = 0.3 \times 7.5 = 2.25 \text{ N}\)

(b) Using the work-energy principle, the work done against friction and the change in potential energy are considered:

Potential energy loss by B:

\(0.5g \sin 30^{\circ} \times 0.6 = 1.5 \text{ J}\)

Applying the work-energy equation to the system:

\(0.5g \sin 30^{\circ} \times 0.6 + 3.5 \times 0.6 = \frac{1}{2} \times 0.8 \times v^2 + 1.1\)

Solving for \(v\):

\(v = 2.5 \text{ m/s}\)

(a) Calculate the initial and final kinetic energy (KE):

\(\text{KE (final)} = \frac{1}{2} \times 1500 \times 20^2 + \frac{1}{2} \times 750 \times 20^2\)

\(\text{KE (initial)} = \frac{1}{2} \times 1500 \times 30^2 + \frac{1}{2} \times 750 \times 30^2\)

Calculate the potential energy (PE) gain:

\(\text{PE gain} = 2250 \times 10 \times 800 \times 0.08\)

Calculate the work done against friction:

\(\text{WD against friction} = 600 \times 800\)

Use the energy equation:

\(\frac{1}{2} \times 2250 \times 30^2 + \text{DF} \times 800 = 600 \times 800 + \frac{1}{2} \times 2250 \times 20^2 + 2250 \times 10 \times 800 \times 0.08\)

Solve for DF:

\(\text{DF} = 1696.875 \text{ N} \approx 1700 \text{ N}\)

(b) Apply Newton's second law:

\(2400 - 600 = 2250a\)

Or, for the car and trailer separately:

\(T - 200 = 750a\) and \(2400 - 400 - T = 1500a\)

Solve for \(a\) and \(T\):

\(T = 800 \text{ N}\) and \(a = 0.8 \text{ m/s}^2\)

(a) The decrease in kinetic energy (KE) is calculated using the formula:

\(\text{Decrease in KE} = \frac{1}{2} \times 4 \times (12^2 - 8^2)\)

\(= \frac{1}{2} \times 4 \times (144 - 64)\)

\(= \frac{1}{2} \times 4 \times 80\)

\(= 160 \text{ J}\)

(b) The potential energy (PE) gained is:

\(\text{PE gained} = 4g \times 10 \sin 30^{\circ}\)

\(= 4 \times 9.8 \times 10 \times 0.5\)

\(= 200 \text{ J}\)

The total work done is the difference between the potential energy gained and the decrease in kinetic energy:

\(\text{Total work done} = 200 - 160\)

\(= 40 \text{ J}\)

(c) When the force is removed, the block decelerates due to gravity. The acceleration \(a\) is given by:

\(-4g \sin 30^{\circ} = 4a\)

\(a = -5 \text{ m/s}^2\)

Using the equation of motion:

\(-10 = 8t - \frac{1}{2} \times 5t^2\)

Solving for \(t\), we find:

\(t = 4.16 \text{ s}\)

(a) To find the speed of P when it has travelled halfway, we first calculate the potential energy (PE) lost. The initial height of P is given by the vertical component of the string length:

\(h = 4 \cos 45^{\circ}\)

When P is halfway, the height is:

\(h = 4 \cos 22.5^{\circ}\)

The PE lost is:

\(\text{PE lost} = 35g(4 \cos 22.5^{\circ} - 4 \cos 45^{\circ})\)

This PE lost is converted into kinetic energy (KE):

\(\frac{1}{2} \times 35 \times v^2 = 35g(4 \cos 22.5^{\circ} - 4 \cos 45^{\circ})\)

Solving for \(v\), we find:

\(v = 4.16 \text{ m/s}\)

(b) Using the work-energy principle, the equation is:

\(\frac{1}{2} \times 35 \times 4^2 = 35g(4 - 4 \cos 45^{\circ}) - X\)

Given the speed at the lowest point is 4 m/s, the KE at the lowest point is:

\(\frac{1}{2} \times 35 \times 4^2\)

Solving for \(X\), we find:

\(X = 130 \text{ J}\)