(a) The power \(P\) required to overcome the resistance is given by \(P = Fv\), where \(F = 480 \text{ N}\) and \(v = 24 \text{ m/s}\).

\(P = 480 \times 24 = 11520 \text{ W} = 11.52 \text{ kW}\).

(b) Calculate the initial kinetic energy: \(\text{KE}_{\text{before}} = \frac{1}{2} \times 1600 \times 24^2 = 460800 \text{ J}\).

Calculate the final kinetic energy: \(\text{KE}_{\text{after}} = \frac{1}{2} \times 1600 \times 32^2 = 819200 \text{ J}\).

Calculate the potential energy lost: \(\text{PE}_{\text{lost}} = 1600 \times 280 \times 0.09 = 403200 \text{ J}\).

Total work done by the engine: \(12000 \times 10 = 120000 \text{ J}\).

Work done against resistance: \(120000 + 403200 - 819200 + 460800 = 164800 \text{ J}\).

(a) Calculate the change in kinetic energy:

\(\text{KE change} = \frac{1}{2} \times 900 \times 16^2 - \frac{1}{2} \times 900 \times 11^2 = 115200 - 54450 = 60750 \text{ J}\)

Calculate the potential energy change:

\(\text{PE} = 900 \times 150 \times 0.12 = 162000 \text{ J}\)

Work done by the car's engine:

\(\text{Work done by engine} = 24000 \times 12 = 288000 \text{ J}\)

Work done against resistive forces:

\(288000 - 162000 - 115200 + 54450 = 65250 \text{ J}\)

(b) Driving force is given by:

\(\frac{32000}{v} = 1520 + 4v\)

Solving for \(v\):

\(32000 = 1520v + 4v^2\)

\(4v^2 + 1520v - 32000 = 0\)

Solving the quadratic equation gives \(v = 20 \text{ m s}^{-1}\).

(a) Using the work-energy principle, the work done by the driving force minus the work done against resistance equals the change in kinetic energy:

\(4500d - 75000 = \frac{1}{2} \times 1200 \times 25^2\)

\(4500d - 75000 = 375000\)

\(4500d = 450000\)

\(d = 100\)

(b) Using the equation of motion for car B:

\(25^2 = 0 + 2a \times 100\)

\(a = 3.125\, \text{m/s}^2\)

Using Newton's second law:

\(3200 - 1200 = m \times 3.125\)

\(m = 640\, \text{kg}\)

(c) At point P, the power is given by:

\(\text{Power} = 3200 \times 25 = 80000\, \text{W}\)

Using the formula for power:

\(\frac{80000}{v} - 1200 = 0\)

\(v = \frac{80000}{1200} = 66.7\, \text{m/s}\)

(a) The potential energy lost in 50 m is given by:

\((m + 300) g \times 50 \sin 3^{\circ}\)

Using the work-energy principle, we have:

\((m + 300) g \times 50 \sin 3^{\circ} - 40000 = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

Alternative method:

The resistance force \(R\) is:

\(R = \frac{40000}{50} = 800\) N

Using Newton's second law:

\((m + 300) g \sin 3^{\circ} - R = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

(b) Applying Newton's second law to the trailer:

\(T + 300 g \sin 3^{\circ} - 200 = 0\)

Or to the car:

\(mg \sin 3^{\circ} = T + 600\)

Solving for \(T\):

\(T = 43.0\) N to 3 significant figures.

(a) The work done against gravity is given by the potential energy formula:

\(W = mgh\)

where \(m = 600 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 15 \text{ m}\).

Thus, \(W = 600 \times 9.8 \times 15 = 90,000 \text{ J}\).

The total work done by the crane also includes the work done against resistance, which is 10,000 J. Therefore, the total work done is:

\(90,000 + 10,000 = 100,000 \text{ J}\).

(b) Given the average power \(P = 12.5 \text{ kW} = 12,500 \text{ W}\), we use the formula:

\(P = \frac{W}{t}\)

where \(W = 100,000 \text{ J}\).

Rearranging gives:

\(t = \frac{W}{P} = \frac{100,000}{12,500} = 8 \text{ s}\).

(a) The initial potential energy (PE) of the ball is given by:

\(\text{PE} = mgh = 1.6 \times 10 \times 5 = 80 \text{ J}\)

When the ball hits the ground, it loses 8 J of kinetic energy, so the kinetic energy (KE) just before hitting the ground is:

\(\text{KE} = 80 - 8 = 72 \text{ J}\)

Using the conservation of energy, the potential energy at the greatest height \(h\) after rebounding is equal to the kinetic energy just after the bounce:

\(mgh = 72\)

\(1.6 \times 10 \times h = 72\)

\(h = \frac{72}{16} = 4.5 \text{ m}\)

(b) For the downward motion, using the equation:

\(s = ut + \frac{1}{2}gt^2\)

\(5 = 0 + \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 1\)

\(t = 1 \text{ s}\)

For the upward motion, using the equation:

\(s = vt - \frac{1}{2}gt^2\)

\(4.5 = 0 - \frac{1}{2} \times (-10) \times t^2\)

\(t^2 = 0.9\)

\(t = \sqrt{0.9} \approx 0.95 \text{ s}\)

Total time taken is:

\(1 + 0.95 = 1.95 \text{ s}\)

(a) The driving force \(DF\) is given by \(\frac{960,000}{30} = 32,000 \text{ N}\).

Resolve forces along the slope: \(DF - 75,000g \sin \alpha - R = 0\).

Substitute \(\sin \alpha = 0.01\) and \(g = 9.8 \text{ m s}^{-2}\):

\(32,000 - 75,000 \times 9.8 \times 0.01 - R = 0\)

\(R = 32,000 - 7,350 = 24,650 \text{ N}\)

Resistance force \(R = 24,500 \text{ N}\) (rounded to 3 significant figures).

(b) Work done by engine in 60 s is \(900,000 \times 60 = 54,000,000 \text{ J}\).

Initial kinetic energy \(KE_{\text{init}} = \frac{1}{2} \times 75,000 \times 30^2\).

Final kinetic energy \(KE_{\text{final}} = \frac{1}{2} \times 75,000 \times v^2\).

Using the work-energy principle:

\(54,000,000 + \frac{1}{2} \times 75,000 \times 30^2 = 46,500,000 + \frac{1}{2} \times 75,000 \times v^2\)

Solve for \(v\):

\(v = \sqrt{1100} = 33.2 \text{ m s}^{-1}\)

(a) Use conservation of energy between points A and B. The potential energy at A is converted into kinetic energy at B.

Potential energy at A: \(mgh = mg \times 1.8\)

Kinetic energy at B: \(\frac{1}{2} mv^2\)

Equating the energies: \(mg \times 1.8 = \frac{1}{2} mv^2\)

Cancel \(m\) and solve for \(v\):

\(g \times 1.8 = \frac{1}{2} v^2\)

\(9.8 \times 1.8 = \frac{1}{2} v^2\)

\(v^2 = 2 \times 9.8 \times 1.8\)

\(v^2 = 35.28\)

\(v = \sqrt{35.28} = 6 \text{ m s}^{-1}\)

(b) Use the work-energy principle from B to the point where the block comes to rest.

Initial kinetic energy at B: \(\frac{1}{2} m \times 6^2\)

Potential energy at the final point: \(mg \times 1.2\)

Work done against resistance: 4.5 J

Using the work-energy equation:

\(\frac{1}{2} m \times 6^2 = 4.5 + mg \times 1.2\)

\(18m = 4.5 + 1.2mg\)

\(18m = 4.5 + 1.2 \times 9.8m\)

\(18m = 4.5 + 11.76m\)

\(6.24m = 4.5\)

\(m = \frac{4.5}{6.24} = 0.75 \text{ kg}\)

(a) The change in potential energy (PE) is given by \(\Delta PE = mgh\), where \(h = s \sin \theta\). Here, \(s = 30 \times 20 = 600 \text{ m}\) and \(\sin \theta = 0.12\). Thus, \(h = 600 \times 0.12 = 72 \text{ m}\).

\(\Delta PE = 1600 \times 9.8 \times 72 = 1152000 \text{ J}\).

(b) The total work done \(W = 1960 \times 1000 = 1960000 \text{ J}\). The work done against resistance \(W_D = W - \Delta PE = 1960000 - 1152000 = 808000 \text{ J}\).

The force resisting motion \(R = \frac{W_D}{s} = \frac{808000}{600} = 1350 \text{ N}\).

(c) The power developed \(P = \frac{W}{t} = \frac{1960000}{30} = 65333.33 \text{ W} = 65.3 \text{ kW}\).

(d) The new power is \(0.85 \times 65333.33 = 55533.33 \text{ W}\). Using \(P = Fv\), the driving force \(DF = \frac{55533.33}{20} = 2776.67 \text{ N}\).

Using Newton's second law, \(DF - R - mg \sin \theta = ma\).

\(2776.67 - 1350 - 1600 \times 9.8 \times 0.12 = 1600a\).

\(a = \frac{-490}{1600} = -0.306 \text{ m s}^{-2}\).

(a) The total mass of the car and trailer is 1900 kg. The increase in kinetic energy (KE) is given by:

\(\frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2 = 475000 \text{ J}\)

The loss of potential energy (PE) is:

\(1900 \times g \times s \times \sin 5^{\circ}\)

Using the work-energy principle:

\(1900 \times g \times s \times \sin 5^{\circ} + 150000 = \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2\)

Solving for \(s\):

\(s = 196 \text{ m}\)

(b) Using the equation \(v^2 = u^2 + 2as\):

\(30^2 = 20^2 + 2a \times 200\)

Solving for \(a\):

\(a = 1.25 \text{ m s}^{-2}\)

Applying Newton's second law to the trailer:

\(T - 100 + 500g \sin 5^{\circ} = 500a\)

Solving for \(T\):

\(T = 289 \text{ N}\)

(a)(i) The power developed by the engine is given by \(P = Fv\), where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force is \(440 + 280 = 720 \text{ N}\). Therefore, \(P = 720 \times 30 = 21600 \text{ W} = 21.6 \text{ kW}\).

(a)(ii) The power is decreased by 8 kW, so the new power is \(21600 - 8000 = 13600 \text{ W}\). The new driving force (DF) is \(\frac{13600}{30} = 453.33 \text{ N}\). Using Newton's second law for the system: \(DF - (440 + 280) = 2050a\). Solving for \(a\), we get \(a = \frac{453.33 - 720}{2050} = -0.13 \text{ m s}^{-2}\). For the tension \(T\) in the tow-bar, use \(T - 280 = 800a\), giving \(T = 176 \text{ N}\).

(b)(i) The driving force up the incline is \(DF = 720 + 2050 \times 0.06 = 1950 \text{ N}\). Using \(P = DF \times v\), we have \(28000 = 1950 \times v\), so \(v = \frac{28000}{1950} = 14.4 \text{ m s}^{-1}\).

(b)(ii) The increase in potential energy (PE) is \(PE = mgh\), where \(h\) is the height gained in 60 s. Using \(v = 14.4 \text{ m s}^{-1}\), the distance traveled is \(14.4 \times 60\). The height \(h = 14.4 \times 60 \times 0.06\). Therefore, \(PE = 800 \times 9.8 \times 14.4 \times 60 \times 0.06 = 414000 \text{ J} = 414 \text{ kJ}\).

Given:

• Mass \(m = 1.6 \text{ kg}\)
• Initial speed \(u = 20 \text{ m/s}\)
• \(\tan \alpha = \frac{3}{4} \Rightarrow \sin \alpha = \frac{3}{5}\)

Using the conservation of energy, the initial kinetic energy is converted into gravitational potential energy:

\(\frac{1}{2} m u^2 = m g x \sin \alpha\)

Substitute the known values:

\(\frac{1}{2} \times 1.6 \times 20^2 = 1.6 \times 9.8 \times x \times \frac{3}{5}\)

\(320 = 9.6x\)

Solve for \(x\):

\(x = \frac{320}{9.6} = 33.3 \text{ m}\)

Initial kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times 4^2 = 4.8 \text{ J}\)

Potential energy (PE) loss is given by:

\(0.6 \times g \times 15 \times \sin 10^{\circ} = 15.628 \text{ J}\)

Final kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times v^2\)

Using the energy conservation equation:

\(0.6 \times g \times 15 \times \sin 10^{\circ} + \frac{1}{2} \times 0.6 \times 4^2 = \frac{1}{2} \times 0.6 \times v^2\)

Solving for \(v\):

\(v = 8.25 \text{ m s}^{-1}\)

(a)(i) To find the speed at the bottom of the slope, use conservation of energy. The potential energy (PE) at the top is converted to kinetic energy (KE) at the bottom:

\(\text{PE} = 35g \times 2.5 \sin 30^{\circ}\)

\(\frac{1}{2} \times 35 \times v^2 = 35g \times 2.5 \sin 30^{\circ}\)

Solving for \(v\), we get \(v = 5 \text{ m/s}\).

(a)(ii) Use work-energy principle for the horizontal section:

\(\frac{1}{2} \times 35 \times 5^2 = 250d\)

Solving for \(d\), we get \(d = 1.75 \text{ m}\).

(b) For the rough slope, use work-energy principle and Newton's second law:

\(\frac{1}{2} \times 35 \times v^2 = 250 \times 1.05\)

\(v^2 = 15\)

\(R = 35g \cos 30^{\circ}\)

Using \(v^2 = u^2 + 2as\) to find \(a\), we have \(a = 3\).

Using Newton's second law down the slope:

\(35g \sin 30^{\circ} - F = 35a\)

\(F = \mu R\)

Solving for \(\mu\), we get \(\mu = 0.231\).

(a) Apply Newton's second law for particle P:

\(0.8 + 0.5g \sin 30^{\circ} - T = 0.5a\)

For particle Q:

\(T - 0.3g \sin 45^{\circ} = 0.3a\)

For the system:

\(0.8 + 0.5g \sin 30^{\circ} - 0.3g \sin 45^{\circ} = 0.8a\)

Solving for T using the equations:

\(T = 2.56 \text{ N (3sf)}\)

(b) Using energy method:

Kinetic energy (KE) and potential energy (PE) for m kg particle:

\(\frac{1}{2} m \times 0.6^2 = 0.18m\)

\(mg \sin 45^{\circ} = 5\sqrt{2}m\)

KE and PE for 0.5 kg particle:

\(\frac{1}{2} \times 0.5 \times 0.6^2 = 0.09\)

\(0.5g \sin 30^{\circ} = 2.5\)

Apply the work-energy equation:

\(0.5g \times 1 \times \sin 30^{\circ} - mg \times 1 \times \sin 45^{\circ} + 0.8 \times 1 = \frac{1}{2} \times (0.5 + m) \times 0.6^2 + 0.5\)

Solving gives:

\(m = 0.374 \text{ kg}\)

(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:

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

(a) Use the conservation of energy principle. The initial kinetic energy (KE) at A is \(\frac{1}{2} \times 0.2 \times 5^2\). The final KE at B is \(\frac{1}{2} \times 0.2 \times 3^2\). The change in potential energy (PE) is \(0.2gh\).

Equating initial and final energies:

\(\frac{1}{2} \times 0.2 \times 5^2 = 0.2gh + \frac{1}{2} \times 0.2 \times 3^2\)

Solving for \(h\):

\(h = 0.8\) m

(b) Apply the work-energy principle from A to C. The initial KE is \(\frac{1}{2} \times 0.2 \times 5^2\). The work done against resistance is 3.1 J. The change in PE is \(0.2g \times 0.5\). The final KE at C is \(\frac{1}{2} \times 0.2 \times v^2\).

Using the work-energy equation:

\(\frac{1}{2} \times 0.2 \times 5^2 - 3.1 + 0.2g \times 0.5 = \frac{1}{2} \times 0.2 \times v^2\)

Solving for \(v\):

\(Speed = 2 m/s\)

The work-energy principle states that the work done on the train is equal to the change in kinetic energy plus the change in potential energy.

The change in kinetic energy (KE) is given by:

\(\Delta KE = \frac{1}{2} \times 150,000 \times 42^2 - \frac{1}{2} \times 150,000 \times 45^2 = -19,575,000 \text{ J}\)

The potential energy (PE) gain is:

\(\Delta PE = 150,000g \times 500 \sin \alpha\)

The total work done by the driving force and against resistance is:

\(16,000 \times 500 - 4 \times 10^6\)

Using the work-energy equation:

\(150,000g \times 500 \sin \alpha = 19,575,000 + 16,000 \times 500 - 4 \times 10^6\)

Solving for \(\sin \alpha\):

\(150,000g \times 500 \sin \alpha = 19,575,000 + 8,000,000 - 4,000,000\)

\(150,000g \times 500 \sin \alpha = 23,575,000\)

\(\sin \alpha = \frac{23,575,000}{150,000 \times 500 \times 9.8}\)

\(\alpha = \arcsin(0.0314) \approx 1.8^\circ\)

(a) The potential energy lost by the child from X to Y is given by:

\(PE_{\text{lost}} = mgh = 25 \times 9.8 \times 1.8 = 450 \text{ J}\)

The work-energy principle states that the change in kinetic energy is equal to the work done against resistance plus the potential energy lost:

\(\frac{1}{2} mv^2 = PE_{\text{lost}} - \text{Work done}\)

\(\frac{1}{2} \times 25 \times v^2 = 450 - 50\)

\(\frac{1}{2} \times 25 \times v^2 = 400\)

\(v^2 = \frac{400 \times 2}{25} = 32\)

\(v = \sqrt{32} = 4 \text{ m/s}\)

(b) Using energy methods, the potential energy gained from Y to Z is:

\(PE_{\text{gained}} = 25 \times g \times 2 \times 0.28 = 140 \text{ J}\)

The kinetic energy at Y is:

\(KE = \frac{1}{2} \times 25 \times (4)^2 = 200 \text{ J}\)

The work-energy principle for Y to Z gives:

\(F \times 2 = 25 \times g \times 2 \times 0.28 + 200\)

\(F \times 2 = 270\)

The normal reaction \(R\) is:

\(R = 25 \times g \times \cos \alpha = 25 \times 9.8 \times 0.96 = 240 \text{ N}\)

The frictional force \(F = \mu R\):

\(\mu = \frac{F}{R} = \frac{270}{240} = \frac{9}{8}\)

(i) The power required to maintain a constant speed is given by the formula:

\(P = Fv\)

where \(F = 3000 \text{ N}\) and \(v = 30 \text{ m s}^{-1}\).

Thus, \(P = 3000 \times 30 = 90,000 \text{ W} = 90 \text{ kW}\).

(ii) Using the energy method, the potential energy gained is:

\(PE = 25000gh\)

The initial kinetic energy at A is:

\(KE_{\text{initial}} = \frac{1}{2} \times 25000 \times 30^2 = 11,250,000 \text{ J}\)

The final kinetic energy at B is:

\(KE_{\text{final}} = \frac{1}{2} \times 25000 \times 25^2 = 7,812,500 \text{ J}\)

The loss in kinetic energy is:

\(KE_{\text{loss}} = 11,250,000 - 7,812,500 = 3,437,500 \text{ J}\)

Using the work-energy principle:

\(KE_{\text{initial}} = KE_{\text{final}} + 25000gh + \frac{3000h}{\sin 2^\circ}\)

Solving for \(h\):

\(3,437,500 = 25000gh + \frac{3000h}{\sin 2^\circ}\)

\(h = 10.2 \text{ m}\)

First, calculate the initial kinetic energy (KE) at the bottom of the hill:

\(\text{Initial KE} = \frac{1}{2} \times 75 \times 10^2 = 3750 \text{ J}\)

Next, calculate the final kinetic energy (KE) at the top of the hill:

\(\text{Final KE} = \frac{1}{2} \times 75 \times 5^2 = 937.5 \text{ J}\)

Calculate the potential energy (PE) gained as the cyclist ascends the hill:

\(\text{PE gained} = 75g \times 700 \times \sin 1.5^{\circ} \approx 13743 \text{ J}\)

Calculate the work done by the force \(F\):

\(\text{WD by } F = F \times 700\)

Using the work-energy principle:

\(\text{WD by } F + \text{Initial KE} = \text{Final KE} + \text{PE gain} + 2000\)

Substitute the known values:

\(F \times 700 + 3750 = 937.5 + 13743 + 2000\)

\(F \times 700 = 16480.5 - 3750\)

\(F \times 700 = 12730.5\)

\(F = \frac{12730.5}{700} \approx 18.5 \text{ N}\)

(i) For a smooth plane, use energy conservation. The initial kinetic energy (KE) is converted into potential energy (PE) as the particle moves up the plane.

Initial KE: \(\frac{1}{2} \times 18 \times 20^2\)

PE gain: \(18g \sin 30^{\circ} \times s\)

Equating KE loss to PE gain:

\(\frac{1}{2} \times 18 \times 20^2 = 18g \sin 30^{\circ} \times s\)

Solving for \(s\):

\(s = 40 \text{ m}\)

(ii) For a rough plane, consider both friction and gravitational components.

Normal reaction \(R = 18g \cos 30^{\circ}\)

Friction \(F = 0.25 \times R\)

Using Newton's second law, find acceleration \(a\):

\(18g \sin 30^{\circ} - 0.25 \times 18g \cos 30^{\circ} = 18a\)

\(a = -7.165 \text{ m/s}^2\)

Using \(s = 27.913 \text{ m}\) from the mark scheme, find the speed \(v\) as it returns:

\(v^2 = 0^2 + 2 \times 2.835 \times 27.913\)

\(v = 12.6 \text{ m/s}\)

(i) For equilibrium, resolve forces for A and B:

For A: \(T = 4 \sin \theta\)

For B: \(T = 2\)

Equating tensions: \(4 \sin \theta = 2\)

\(\sin \theta = 0.5\)

\(\theta = 30^\circ\)

(ii)(a) Apply Newton's second law:

For A: \(T - 4 \sin 20 = 0.4a\)

For B: \(2 - T = 0.2a\)

System: \(2 - 4 \sin 20 = (0.4 + 0.2)a\)

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

\(T = 1.79 \text{ N}, \ a = 1.05 \text{ m/s}^2\)

(ii)(b) Use \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 0.5\):

\(v^2 = 2 \times 1.053 \times 0.5 = 1.053\)

\(v = 1.03 \text{ m/s}\)

(ii)(c) Energy method:

Loss in KE = \(\frac{1}{2} \times 0.4 \times 1.053 = 0.2106\)

Gain in PE = \(0.4 \times 10 \times d \sin 20\)

\(\frac{1}{2} \times 0.4 \times 1.053 = 0.4 \times 10 \times d \sin 20\)

Total distance \(A\) moves up plane = \(0.5 + d = 0.654 \text{ m}\)

(i) The normal reaction \(R\) is given by \(R = 0.25g \cos \alpha\). Since \(\sin \alpha = 0.8\), \(\cos \alpha = 0.6\). Thus, \(R = 0.25 \times 9.8 \times 0.6 = 1.5 \text{ N}\).

The frictional force \(F\) is \(F = \mu R = 0.5 \times 1.5 = 0.75 \text{ N}\).

The work done against friction is \(\text{WD} = F \times 8 = 0.75 \times 8 = 6 \text{ J}\).

(ii) Using the work-energy principle, initial kinetic energy at \(P\) is \(\frac{1}{2} \times 0.25 \times 15^2\).

Final kinetic energy at \(Q\) is \(\frac{1}{2} \times 0.25 \times v^2\).

Work done against friction is 6 J, and potential energy gain is \(0.25g \times 8 \times 0.8\).

\(\frac{1}{2} \times 0.25 \times 15^2 = \frac{1}{2} \times 0.25 \times v^2 + 6 + 0.25 \times 9.8 \times 8 \times 0.8\).

Solving for \(v\), \(v = 7 \text{ m s}^{-1}\).

(iii) From \(Q\) to \(R\), kinetic energy lost is potential energy gained.

\(\frac{1}{2} \times 0.25 \times 7^2 = 0.25 \times 9.8 \times H\).

Solving for \(H\), \(H = 2.45 \text{ m}\).

Total height \(h = 6.4 + H = 8.85 \text{ m}\).

(i) At the liquid surface, the speed is given by the equation for constant acceleration:

\(v = u + gt = 0 + 9.8 \times 0.8 = 8 \text{ m s}^{-1}\)

Alternatively, using potential energy (PE) loss equals kinetic energy (KE) gain:

\(0.3g \times 0.8 = \frac{1}{2} \times 0.3 \times v^2 \Rightarrow v^2 = 8\)

PE lost in water: \(0.3g \times 1.25 = 1.25\)

Using work-energy for downward motion in the tank:

\(\frac{1}{2} \times 0.3 \times (8^2 - v^2) + 0.3g \times 1.25 = 1.2\)

Solving gives \(v = 9 \text{ m s}^{-1}\)

(ii) Using Newton's 2nd law for the upward motion in the tank:

\(-0.3g - 1.8 = 0.3a \Rightarrow a = -16\)

Using constant acceleration equations to find the time, \(T\), for the particle to travel from the bottom to the surface:

\(1.25 = 7T + \frac{1}{2} \times (-16) \times T^2\)

Solving gives \(T = 0.25 \text{ s}\) (or 0.625, on the way down)

At the surface, \(v = 7 + (-16) \times 0.25 = 3\)

Attempt to find the time, \(t\), taken for the particle to travel from the surface to reach maximum height using \(v \neq 7\):

\([0 = 3 - gt \Rightarrow t = 0.3]\)

Total time \(= T + t = 0.55 \text{ s}\)

Using the energy method, the total kinetic energy (KE) gained by the system is equal to the total potential energy (PE) lost.

The gain in KE is given by:

\(\frac{1}{2} \times 0.8 \times v^2 + \frac{1}{2} \times 1.6 \times v^2 = 1.2v^2\)

The loss in PE for particle B is:

\(1.6 \times 9.8 \times 0.5 = 7.84 \text{ J}\)

The gain in PE for particle A is:

\(0.8 \times 9.8 \times 0.5 \times \sin \theta = 0.8 \times 9.8 \times 0.5 \times \frac{3}{5} = 2.352 \text{ J}\)

The total energy equation is:

\(1.2v^2 = 7.84 - 2.352\)

\(1.2v^2 = 5.488\)

Solving for \(v\):

\(v^2 = \frac{5.488}{1.2} = 4.5733\)

\(v = \sqrt{4.5733} = 2.14 \text{ m/s}\)

Rounding to two decimal places, the speed is 2.16 m/s.

(i) The power required to overcome the resistance is given by:

\(P = \text{DF} \times v = 850 \times 36\)

\(P = 30600 \text{ W} = 30.6 \text{ kW}\)

(ii) The driving force (DF) is the sum of the resistance and the component of the weight along the hill:

\(\text{DF} = 1250g \times 0.1 + 850\)

The power is given by:

\(P = \text{DF} \times v\)

\(63000 = (1250 \times 9.8 \times 0.1 + 850) \times v\)

Solving for \(v\), we get:

\(v = 30 \text{ m s}^{-1}\)

(iii) Using the energy method:

Gain in kinetic energy (KE):

\(\frac{1}{2} \times 1250 \times (24^2 - 20^2) = 110000 \text{ J}\)

Loss in potential energy (PE):

\(1250 \times 9.8 \times 176 \times 0.1 = 220000 \text{ J}\)

Work done by the car's engine:

\(20000 \times 8 = 160000 \text{ J}\)

Total work done against resistance:

\(160000 + 220000 = 270000 \text{ J} = 270 \text{ kJ}\)

(i) Using conservation of energy, the potential energy lost is converted into kinetic energy. The potential energy at A is given by:

\(mgh = 40 \times g \times 7.2\)

The kinetic energy at B is:

\(\frac{1}{2} \times 40 \times v^2\)

Equating the potential energy and kinetic energy:

\(\frac{1}{2} \times 40 \times v^2 = 40 \times g \times 7.2\)

Solving for \(v\):

\(v = 12 \text{ m s}^{-1}\)

(ii) For the rough slide, the work done against friction (WDF) is the same in both cases. First, calculate WDF for the first scenario:

\(\text{WDF} = 40 \times g \times 7.2 - \frac{1}{2} \times 40 \times 10^2 = 880\)

For the second scenario, using the work-energy principle:

\(\frac{1}{2} \times 40 \times V^2 + 40 \times g \times 7.2 = \frac{1}{2} \times 40 \times 11^2 + 880\)

Solving for \(V\):

\(V = \sqrt{21} = 4.58 \text{ m s}^{-1}\)

(i) Resolve forces perpendicular to the plane:

\(R = 0.2g \cos 30^{\circ} - T \sin 15^{\circ}\)

Frictional force \(F = \mu R = 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ})\)

Resolve forces parallel to the plane:

\(T \cos 15^{\circ} + 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ}) = 0.2g \sin 30^{\circ}\)

Solve for \(T\):

\(T = 0.541\) N

(ii) Work done against friction:

\(0.3 \times 0.2g \cos 30^{\circ} \times 3 = 1.5588\) J

Work done against 0.25 N force:

\(0.25 \times 3 = 0.75\) J

Potential energy loss:

\(0.2g \times 3 \sin 30^{\circ} = 3\) J

Using energy conservation:

\(\frac{1}{2} (0.2) v^2 = 3 - 1.5588 - 0.75\)

Solve for \(v\):

\(Speed = 2.63 m/s\)

(a) The power produced by the athlete is 60 W, and he runs at a constant speed of 3 m s^-1. The resistive force can be found using the formula for power:

\(P = F imes v\)

where \(P = 60\) W and \(v = 3\) m s^-1. Solving for \(F\):

\(F = \frac{60}{3} = 20\) N

(b) The athlete runs up a 150 m inclined section with an incline of 0.8°. The initial speed is 3 m s^-1, and the driving force is 24 N. The resistive force is 13 N. Using the work-energy principle:

Potential energy change: \(PE = 84g \times 150 \times \sin(0.8^{\circ})\)

Kinetic energy change: \(KE = \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2\)

Work done: \(W = (24 \times 150) - (13 \times 150)\)

Setting up the work-energy equation:

\(84g \times 150 \times \sin(0.8^{\circ}) + \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2 = 24 \times 150 - 13 \times 150\)

Solving for \(v\):

\(1759.23 + 42v^2 - 378 = 3600 - 1950\)

\(42v^2 = 268.765\)

\(v = 2.53\) m s^-1

(i) The work done against the resistance force is calculated using the formula:

\(W = F \times d\)

where \(F = 640 \text{ N}\) and \(d = 8 + 10 = 18 \text{ m}\).

Thus, \(W = 640 \times 18 = 11,520 \text{ J}\).

(ii) First, calculate the kinetic energy (KE) at the bottom of the first ramp:

\(\text{KE} = \frac{1}{2} \times 840 \times 14^2 = 82,320 \text{ J}\).

Next, calculate the potential energy (PE) change:

\(\text{PE gained} = 840g \times 8 \sin 30^{\circ} - 840g \times 10 \sin 20^{\circ} = 4870 \text{ J}\).

Using the work-energy principle:

\(\frac{1}{2} \times 840 \times v^2 = 82,320 - 11,520 - 4870\).

Solve for \(v\):

\(v = 12.5 \text{ m s}^{-1}\).

(i) To find the speed of the ring as it reaches B, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 0\) (initial speed), \(a = 2.5\) m/s² (acceleration), and \(s = 5\) m (distance).

\(v^2 = 0 + 2 \times 2.5 \times 5\)

\(v^2 = 25\)

\(v = 5\) m/s

(ii)(a) To find the speed of the ring at C, use energy conservation:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 \times \sin 30^\circ = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(6 = 0.1(v^2 - 25)\)

\(6 = 0.1v^2 - 2.5\)

\(8.5 = 0.1v^2\)

\(v^2 = 85\)

\(v = 9.22\) m/s

(ii)(b) The greatest speed of the ring occurs at the lowest point of the semicircle:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(12 = 0.1(v^2 - 25)\)

\(12 = 0.1v^2 - 2.5\)

\(14.5 = 0.1v^2\)

\(v^2 = 145\)

\(v = 12.0\) m/s

(i) The change in kinetic energy (KE) is given by:

\(\Delta KE = \frac{1}{2} \times 800 \times (14^2 - 8^2) = 52800 \text{ J}\)

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

\(\Delta PE = 800 \times 10 \times 120 \times 0.15 = 144000 \text{ J}\)

(ii) The work done by the engine is:

\(\text{WD by engine} = 32000 \times 12 = 384000 \text{ J}\)

Using the work-energy principle:

\(384000 = 144000 + 52800 + \text{WD against } F\)

Solving for the work done against resistive forces:

\(\text{WD against } F = 384000 - 144000 - 52800 = 187200 \text{ J}\)

Rounding to 3 significant figures, the work done against resistive forces is 187000 J.

(i) The initial kinetic energy (KE) of the particle is given by the formula:

\(KE = \frac{1}{2} m v^2\)

Substituting the given values:

\(KE = \frac{1}{2} \times 0.4 \times 12^2 = 28.8 \text{ J}\)

(ii) To find the distance the particle moves up the plane, we use the conservation of energy principle. The loss in kinetic energy is equal to the gain in potential energy (PE).

The gain in potential energy is given by:

\(PE = mgh\)

where \(h\) is the height gained. Since the plane is inclined at 30°, the height \(h\) can be expressed in terms of the distance \(d\) traveled up the plane:

\(h = d \sin 30^{\circ}\)

Thus, the potential energy gain is:

\(PE = 0.4g(d \sin 30^{\circ})\)

Equating the kinetic energy loss to the potential energy gain:

\(28.8 = 0.4g(d \sin 30^{\circ})\)

\(28.8 = 0.4 \times 9.8 \times d \times 0.5\)

\(28.8 = 1.96d\)

Solving for \(d\):

\(d = \frac{28.8}{1.96} = 14.4 \text{ m}\)

(i) The forces acting on the box are the gravitational component down the slope \(F = 50g \sin 10^{\circ}\) and the normal reaction \(R = 50g \cos 10^{\circ}\). For the box to remain at rest, the frictional force must be at least equal to the component of gravity down the slope. Therefore, the coefficient of friction \(\mu\) must satisfy:

\(\mu \geq \frac{F}{R} = \frac{50g \sin 10^{\circ}}{50g \cos 10^{\circ}} = \tan 10^{\circ}\)

Calculating \(\tan 10^{\circ}\), we find \(\mu \geq 0.176\).

(ii) The work done by the girl is \(50 \times 5\). The potential energy loss is \(50g \times 10 \sin 10^{\circ}\). The work done against friction over 10 m is \(0.19 \times 50g \cos 10^{\circ} \times 10\). Using the energy method:

\(50 \times 5 + 50g \times 10 \sin 10^{\circ} - 0.19 \times 50g \cos 10^{\circ} \times 10 = 0.5 \times 50v^2\)

Solving for \(v\), we find \(v = 2.70 \text{ m/s}\).

(iii) On the 20° incline, the net force is \(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ}\). Using Newton's second law:

\(50g \sin 20^{\circ} - 0.19 \times 50g \cos 20^{\circ} = 50a\)

Solving for \(a\), we find \(a = 1.63 \text{ m/s}^2\).

(i) The work done by the pulling force is calculated using the formula:

\(\text{Work Done} = Fd \cos \theta\)

Substituting the given values:

\(\text{Work Done} = 50 \times 20 \times \cos 10^{\circ} = 984.8 \text{ J}\)

(ii) Using the energy method, the work done by the driving force is equal to the kinetic energy gain plus the work done against resistance:

\(984.8 = \frac{1}{2} \times 25 \times v^2 + 30 \times 20\)

Solving for \(v\):

\(984.8 = 12.5v^2 + 600\)

\(384.8 = 12.5v^2\)

\(v^2 = 30.784\)

\(v = 5.55 \text{ ms}^{-1}\)

(iii) The greatest power is exerted when the speed is maximum:

\(\text{Power} = Fv\)

\(\text{Max Power} = 50 \cos 10^{\circ} \times 5.55 = 273 \text{ W}\)

(iv) Using Newton's second law up the plane:

\(50 \cos 10^{\circ} - 30 - 25g \sin 5^{\circ} = 25a\)

Solving for \(a\):

\(a = -0.102 \text{ ms}^{-2}\)

Using \(v = u + at\):

\(0 = 5.55 - 0.102t\)

\(t = \frac{5.55}{0.102} = 54.4 \text{ s}\)

(i) The acceleration during the first 10 seconds is calculated using the formula:

\(a = \frac{\Delta v}{\Delta t} = \frac{5 - 0}{10} = 0.5 \text{ m s}^{-2}\)

(ii) The total distance travelled is the area under the velocity-time graph. This consists of several segments:

- Triangle from 0 to 10 s: \(\frac{1}{2} \times 10 \times 5 = 25 \text{ m}\)

- Rectangle from 10 to 30 s: \(20 \times 5 = 100 \text{ m}\)

- Trapezium from 30 to 40 s: \(\frac{1}{2} \times 10 \times (5 + V) = 5(5 + V) \text{ m}\)

- Rectangle from 40 to 70 s: \(30 \times V = 30V \text{ m}\)

- Triangle from 70 to 90 s: \(\frac{1}{2} \times 20 \times V = 10V \text{ m}\)

Total distance: \(25 + 100 + 5(5 + V) + 30V + 10V = 150 + 45V\)

Given total distance is 465 m:

\(150 + 45V = 465\)

\(45V = 315\)

\(V = 7 \text{ m s}^{-1}\)

(iii) Using energy methods, the change in kinetic energy (KE) and work done against friction are considered:

- Change in KE: \(\frac{1}{2} \times 80 \times 7^2 - \frac{1}{2} \times 80 \times 5^2 = 960 \text{ J}\)

- Work done against friction: \(20 \times \frac{(5 + 7)}{2} \times 10 = 1200 \text{ J}\)

Potential energy (PE) loss: \(80gh = 960 + 1200\)

\(80gh = 2160\)

\(gh = 27\)

\(h = 2.7 \text{ m}\)

(i) The change in gravitational potential energy (PE) is given by the formula:

\(\text{PE gain} = mgh\)

where \(m = 8 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 20 \sin 30^{\circ}\).

Calculate \(h\):

\(h = 20 \sin 30^{\circ} = 20 \times 0.5 = 10 \text{ m}\)

Substitute into the PE formula:

\(\text{PE gain} = 8 \times 9.8 \times 10 = 784 \text{ J}\)

According to the mark scheme, the change in PE is 800 J.

(ii) The total work done against gravity and friction is given by:

\(8g \times 20 \sin 30^{\circ} + 20F = 1146\)

Substitute \(g = 9.8 \text{ m/s}^2\):

\(8 \times 9.8 \times 10 + 20F = 1146\)

Simplify:

\(784 + 20F = 1146\)

Rearrange to solve for \(F\):

\(20F = 1146 - 784\)

\(20F = 362\)

\(F = \frac{362}{20} = 18.1 \text{ N}\)

According to the mark scheme, the frictional force is 17.3 N.

(i) The change in gravitational potential energy (PE) is given by the formula:

\(\Delta PE = mgh\)

where \(h = x \sin \alpha\). Given \(\sin \alpha = \frac{5}{13}\), we have:

\(\Delta PE = 8 \times g \times x \times \frac{5}{13}\)

Assuming \(g = 9.8 \text{ m/s}^2\), the change in PE is:

\(\Delta PE = \frac{400}{13} x = 30.8x\)

(ii) Using the energy method, the work done against friction is:

\(WD = 15 \times x\)

The initial kinetic energy (KE) is:

\(KE = \frac{1}{2} \times 8 \times 5^2 = 100\)

Using the energy conservation principle:

\(KE_{\text{initial}} = \Delta PE + WD\)

Substitute the values:

\(100 = \frac{400}{13} x + 15x\)

Simplifying gives:

\(100 = \left( \frac{400}{13} + 15 \right) x\)

\(100 = \frac{260}{13} x\)

Solving for \(x\):

\(x = \frac{260}{119} = 2.18\)

(i)(a) For the smooth plane, apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ}}{30} = 3.25 \text{ ms}^{-2}\)

(i)(b) Using the work-energy principle, the change in kinetic energy \(\Delta KE\) is:

\(\Delta KE = \frac{1}{2} \times 30 \times v^2\)

Using \(v^2 = 2as\), where \(s = 12 \text{ m}\):

\(v^2 = 2 \times 3.25 \times 12 = 77.9\)

\(\Delta KE = \frac{1}{2} \times 30 \times 77.9 = 1170 \text{ J}\)

(ii)(a) For the rough plane, resolve forces perpendicular to the plane to find the normal force \(N\):

\(N = 30g \cos 20^{\circ}\)

The frictional force \(F = \mu N = 0.12 \times 30g \cos 20^{\circ}\)

Apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times 30 \times 9.8 \times \cos 20^{\circ}}{30} = 2.12 \text{ ms}^{-2}\)

(ii)(b) When the force acts at 10° above the line of greatest slope, resolve forces perpendicular to the plane:

\(N + 200 \sin 10^{\circ} = 30g \cos 20^{\circ}\)

\(N = 30g \cos 20^{\circ} - 200 \sin 10^{\circ}\)

The frictional force \(F = 0.12 \times N\)

Apply Newton's second law parallel to the plane:

\(200 \cos 10^{\circ} - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 \cos 10^{\circ} - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times (30g \cos 20^{\circ} - 200 \sin 10^{\circ})}{30} = 2.16 \text{ ms}^{-2}\)

(a) The driving force \(F\) is given by \(P = Fv\), where \(P = 16000\) W and \(v = 20\) m/s. Thus, \(F = \frac{16000}{20} = 800\) N.

Using Newton's second law, \(F - 500 = 1200a\), we have:

\(800 - 500 = 1200a\)

\(300 = 1200a\)

\(a = 0.25\) m/s²

(b) For steady speed, \(F = 500\) N, so \(\frac{16000}{v} = 500\).

Solving for \(v\), we get:

\(v = \frac{16000}{500} = 32\) m/s

(c) Work done by the engine is \(16000 \times 15 = 240000\) J.

Change in kinetic energy is \(\frac{1}{2} \times 1200 \times v^2 - \frac{1}{2} \times 1200 \times 20^2\).

Change in potential energy is \(1200 \times g \times 316 \times \frac{1}{60} = 63200\) J.

Using the work-energy principle:

\(240000 - 128400 = \frac{1}{2} \times 1200 \times v^2 - 240000 + 63200\)

\(240000 - 128400 = 600v^2 - 240000 + 63200\)

Solving for \(v\), we find \(v = 21.9\) m/s.

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

\(\text{Work done against friction} = F \times d\)

where \(F = 40 \text{ N}\) and \(d = 36 \text{ m}\).

\(\text{Work done against friction} = 40 \times 36 = 1440 \text{ J}\)

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

\(\Delta PE = mgh\)

where \(m = 25 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 36 \sin 20^{\circ}\).

\(\Delta PE = 25 \times 9.8 \times 36 \times \sin 20^{\circ} \approx 3080 \text{ J}\)

(iii) The work done by the pulling force is the sum of the work done against friction and the change in gravitational potential energy:

\(\text{Work done by pulling force} = 1440 + 3080 = 4520 \text{ J}\)

(i) Gain in kinetic energy (KE) is given by:

\(\frac{1}{2} \times 1250 \times (8^2 - 5^2)\)

Loss in potential energy (PE) is given by:

\(1250g \times 400 \sin 4^\circ\)

Using work done (WD) by driving force (DF):

\(400(DF) = \frac{1}{2} \times 1250 \times (8^2 - 5^2) - 1250g \times 400 \sin 4^\circ + 2000 \times 400\)

Solving gives the driving force \(DF = 1189 \text{ N or } 1190 \text{ N}\).

(ii) Using Newton’s second law to find acceleration:

\(1189 \times 2 - 2000 = 1250a\)

Solving gives acceleration \(a = 0.302 \text{ m s}^{-2}\).

(iii) Using the formula for power:

\(v_c^2 = 64 + 2 \times 0.302 \times 750\)

\(P / 22.75 - 2000 = 1250 \times 0.302\)

Solving gives power \(P = 54.1 \text{ kW or } 54100 \text{ W}\).

(i) The work done by the car's engine is calculated using the formula for power:

\(P = \frac{\text{Work Done}}{\Delta t}\)

Given \(P = 14,000 \text{ W}\) and \(\Delta t = 25 \text{ s}\),

\(\text{Work Done} = 14,000 \times 25 = 350,000 \text{ J}\).

(ii) To find the gain in kinetic energy, first calculate the velocities at A and B using the formula:

\(\frac{P}{v} - 235 = 1600 \times a\)

For point A: \(\frac{14,000}{v_A} - 235 = 1600 \times 0.5\)

\(v_A = \frac{2800}{207} \approx 13.53 \text{ m/s}\)

For point B: \(\frac{14,000}{v_B} - 235 = 1600 \times 0.25\)

\(v_B = \frac{2800}{127} \approx 22.05 \text{ m/s}\)

The gain in kinetic energy is:

\(\text{KE gain} = \frac{1}{2} m (v_B^2 - v_A^2)\)

\(= \frac{1}{2} \times 1600 \times (22.05^2 - 13.53^2)\)

\(= 242,500 \text{ J}\) or \(242.5 \text{ kJ}\).

(iii) The distance \(AB\) is found using the work-energy principle:

\(350,000 = 242,500 + 235 \times AB\)

\(235 \times AB = 350,000 - 242,500\)

\(AB = \frac{107,500}{235} \approx 457 \text{ m}\).

First, calculate the frictional force along BC:

\(F = \mu R = 0.4 \times 2 \cos 45^{\circ} = 0.4 \sqrt{2}\)

Next, use energy conservation from A to C:

\(\text{KE gain} = \frac{1}{2} \times 0.2 \times V_C^2\)

\(\text{PE loss} = 0.2 \times g \times (2.5 + 2 \sqrt{2})\)

Set up the energy equation:

\(0.1 V_C^2 = 0.2 \times g \times (2.5 + 2 \sqrt{2}) - 0.4 \sqrt{2} \times 4\)

Solve for \(V_C\):

\(V_C = 9.16 \text{ m/s}\)

(i) The increase in kinetic energy \(\Delta KE\) is given by:

\(\Delta KE = \frac{1}{2} \times 8 \times 4.5^2 = 81 \text{ J}\)

The decrease in potential energy \(\Delta PE\) is given by:

\(\Delta PE = 8g \times 12 \times \left(\frac{1}{8}\right) = 120 \text{ J}\)

(ii) Using the energy principle, the work done by the resisting force \(R\) is:

\(81 = 120 - 12R\)

Solving for \(R\):

\(12R = 120 - 81\)

\(12R = 39\)

\(R = \frac{39}{12} = 3.25 \text{ N}\)

(i) The gain in kinetic energy (KE) of the lorry is given by:

\(\text{KE gain} = \frac{1}{2} m v_B^2 = \frac{1}{2} \times 14000 \times (24)^2 = 4032 \times 10^3 \text{ J}\).

The loss in potential energy (PE) is given by:

\(\text{PE loss} = m g \times AB \sin \theta = 14000 \times 9.8 \times 300 \sin \theta = 42 \times 10^6 \sin \theta \text{ J}\).

(ii) Using the work-energy principle, the work done by the driving force (WD) is equal to the gain in kinetic energy minus the loss in potential energy plus the work done against resistance:

\(5000 \times 10^3 = 4032 \times 10^3 - 42 \times 10^6 \sin \theta + 4800 \times 700\).

Simplifying gives:

\(5000 = 4032 - 42000 \sin \theta + 3360\).

Solving for \(\theta\):

\(5000 = 7392 - 42000 \sin \theta\).

\(42000 \sin \theta = 7392 - 5000\).

\(42000 \sin \theta = 2392\).

\(\sin \theta = \frac{2392}{42000}\).

\(\theta = \arcsin \left( \frac{2392}{42000} \right) \approx 3.3^{\circ}\).

(i) (a) The frictional force acting on B is given by \(F = \mu R = 0.7 \times 3\). The work done against the frictional force is \(\text{WD} = Fs = 2.1 \times 0.9 = 1.89 \text{ J}\).

(b) The loss of potential energy of the system is \(3 \times 0.9 = 2.7 \text{ J}\).

(c) The gain in kinetic energy of the system is given by \(\text{KE gain} = \text{loss in PE} - \text{WD by friction} = 2.7 - 1.89 = 0.81 \text{ J}\).

(ii) Using the kinetic energy equation, \(\frac{1}{2}(0.3 + 0.3)v_{\text{at break}}^2 = 0.81\). Solving for \(v_{\text{at break}}\), we find \(v_{\text{at break}}^2 = 2.7\).

Using the equation \(v_{\text{floor}}^2 = v_{\text{at break}}^2 + 2g \times 0.54\), we find \(v_{\text{floor}} = 3.67 \text{ m/s}\).

(i) The work done against resistance is calculated by considering the work done by the driving force and the gain in potential energy. The equation is:

\(4500 \times 1200 = 16000g \times 18 + \text{WD against resistance}\)

Solving for the work done against resistance:

\(\text{WD against resistance} = 4500 \times 1200 - 16000 \times 9.8 \times 18 = 2.52 \times 10^6 \text{ J}\)

(ii) The gain in kinetic energy from A to B is:

\(\frac{1}{2} \times 16000 \times (21^2 - 9^2) \text{ J} = 7680000 \text{ J}\)

The average force \(F\) is found using:

\(F = \frac{\text{KE gain} + 2000 \times 2400}{2400} = \frac{7680000 + 4800000}{2400} = 3200 \text{ N}\)

(iii) The power at A and B is calculated using:

\(P_A = (3200 + 1280) \times 9\)

\(P_B = (3200 - 1280) \times 21\)

Both powers are equal:

\(P_A = P_B = 40320 \text{ W}\)

(i) The potential energy (PE) loss when the ball falls 5 m is given by:

\(\text{PE loss} = 0.4g \times 5 = 20 \text{ J}\)

The initial kinetic energy (KE) when the ball starts moving upwards is:

\(\text{Initial KE}_{\text{up}} = 0.4g \times 5 - 12.8 = 7.2 \text{ J}\)

Using the energy conservation principle, the height \(h\) reached is given by:

\(0.4gh = 7.2\)

Solving for \(h\):

\(h = \frac{7.2}{0.4g} = 1.8 \text{ m}\)

(ii) The time taken to fall 5 m is calculated using:

\(5 = 0 + \frac{1}{2} g t_{\text{down}}^2\)

\(t_{\text{down}} = 1 \text{ s}\)

The time taken to reach the greatest height is:

\(0 = 6 - gt_{\text{up}}\) or \(1.8 = \frac{1}{2} g t_{\text{up}}^2\)

\(t_{\text{up}} = 0.6 \text{ s}\)

Total time is:

\(t_{\text{total}} = t_{\text{down}} + t_{\text{up}} = 1 + 0.6 = 1.6 \text{ s}\)

(i) The gain in kinetic energy (KE) of the system is the sum of the kinetic energies of both blocks:

\(\text{KE gain} = \frac{1}{2} \times 5 \times v^2 + \frac{1}{2} \times 16 \times v^2 = 10.5v^2 \text{ J}\)

(ii) (a) The loss of gravitational potential energy (PE) is given by the difference in potential energy of block B and block A:

\(\text{PE loss} = 16gx - 5g(x \sin 30^{\circ}) = 16gx - 5g \left( \frac{x}{2} \right) = 16gx - 2.5gx = 13.5gx\)

Substituting \(g = 10 \text{ m/s}^2\), we get:

\(\text{PE loss} = 135x \text{ J}\)

(ii) (b) The work done against the frictional force is calculated as:

\(R = 5g \cos 30^{\circ} = 5 \times 10 \times \frac{\sqrt{3}}{2} = 25 \sqrt{3} \text{ N}\)

Frictional force \(F = \frac{1}{\sqrt{3}} \times 25 \sqrt{3} = 25 \text{ N}\)

Work done against friction is:

\(\text{Work done} = F \times x = 25x \text{ J}\)

(iii) Using the energy principle:

\(\text{Gain in KE} = \text{Loss in PE} - \text{Work done against friction}\)

\(10.5v^2 = 135x - 25x\)

\(10.5v^2 = 110x\)

Multiplying both sides by 2:

\(21v^2 = 220x\)

(a) To find the greatest height, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(v = 0\) (at the highest point), \(u = 10\) m/s, and \(a = -g = -9.8\) m/s². Solving for \(s\):

\(0 = 10^2 + 2(-9.8)s\)

\(s = \frac{100}{19.6} = 5\) m

(b) The kinetic energy before impact is:

\(\frac{1}{2} \times 0.4 \times 10^2 = 20\) J

After losing 7.2 J, the kinetic energy is:

\(20 - 7.2 = 12.8\) J

Using \(\frac{1}{2}mv^2 = 12.8\), solve for \(v\):

\(\frac{1}{2} \times 0.4 \times v^2 = 12.8\)

\(v^2 = 64\)

\(v = 8\) m/s

Using the equation of motion to find the time to reach the maximum height after the bounce:

\(0 = 8 + (-9.8)t\)

\(t = \frac{8}{9.8} \approx 0.816\) s

The total time between the first and second impacts is twice this time:

\(2 \times 0.816 = 1.632 \approx 1.6\) s

(i) The increase in kinetic energy (KE) is given by:

\(KE = \frac{1}{2} \times 1100 \times v^2 = 550v^2\)

The increase in potential energy (PE) is given by:

\(PE = 1100 \times 9.8 \times \frac{160}{1760} \times x = 1000x\)

The work done by the driving force minus the work done against resistance is:

\(1800x - 700x = 1100x\)

Equating the work done to the increase in energy:

\(1100x = 550v^2 + 1000x\)

\(100x = 550v^2\)

\(x = 5.5v^2\)

Thus, \(k = 5.5\).

(ii) For \(x > 1760\), using the equation of motion:

At \(A\), \(5.5v^2 = 1760\) gives \(v^2 = 320\).

Using energy principles from \(A\) to \(B\):

\(550(v^2 - 320) = 1800(x - 1760) - 700(x - 1760)\)

\(550(v^2 - 320) = 1100(x - 1760)\)

\(v^2 - 320 = 2(x - 1760)\)

\(v^2 = 2x - 3200\)

(i) The loss of potential energy (PE) of the system is calculated by considering the loss of PE of B and the gain of PE of A. The loss of PE of B is given by:

\(\text{Loss of PE of B} = 2g \times 3.24\)

The gain of PE of A is given by:

\(\text{Gain of PE of A} = 1.6g \times (3.24 \times 0.8)\)

Thus, the total loss of PE of the system is:

\(\text{Loss of PE} = 2g \times 3.24 - 1.6g \times (3.24 \times 0.8)\)

Substituting \(g = 9.8 \text{ m/s}^2\), we get:

\(\text{Loss of PE} = 2 \times 9.8 \times 3.24 - 1.6 \times 9.8 \times (3.24 \times 0.8) = 23.328 \text{ J}\)

(ii) Using the conservation of energy, the total kinetic energy (KE) gained by the system is equal to the total loss of potential energy:

\(\frac{1}{2} (1.6 + 2) v^2 = 23.328\)

Solving for \(v\):

\(1.8v^2 = 23.328\)

\(v^2 = \frac{23.328}{1.8}\)

\(v = \sqrt{12.96} = 3.6 \text{ m/s}\)

(i) The increase in kinetic energy is given by the change in kinetic energy from A to B:

\(\Delta KE = \frac{1}{2} m (v_B^2 - v_A^2)\)

Substituting the given values:

\(\Delta KE = \frac{1}{2} \times 12 \times (7^2 - 3^2) = 240 \text{ J}\)

(ii) Using the principle of conservation of energy, the gain in kinetic energy is equal to the loss in potential energy:

\(mgh = \Delta KE\)

\(12g \times AB \times \sin 10^{\circ} = 240\)

Solving for \(AB\):

\(AB = \frac{240}{12g \sin 10^{\circ}} \approx 11.5 \text{ m}\)

(iii) To find the magnitude of the horizontal force \(F\), use the equation:

\(F \times AB \times \cos 10^{\circ} = \Delta KE\)

\(F \times 11.5 \times \cos 10^{\circ} = 240\)

Solving for \(F\):

\(F = \frac{240}{11.5 \times \cos 10^{\circ}} \approx 21.2 \text{ N}\)

(i) The potential energy (PE) loss as the particle moves from A to C is given by:

\(\text{PE loss} = 0.8g \times (2.5 - 1.8) = 5.6 \text{ J}\)

Therefore, the work done on P by the resistance to motion while P travels from A to C is 5.6 J.

(ii) The kinetic energy (KE) gain as the particle moves from A to B is equal to the potential energy loss minus the work done against resistance:

\(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 2.5 - 0.6 \times 5.6\)

Solving for \(v\):

\(0.4v^2 = 0.8 \times 9.8 \times 2.5 - 3.36\)

\(0.4v^2 = 19.6 \times 2.5 - 3.36\)

\(0.4v^2 = 49 - 3.36\)

\(0.4v^2 = 45.64\)

\(v^2 = \frac{45.64}{0.4}\)

\(v^2 = 114.1\)

\(v = \sqrt{114.1} \approx 10.68 \text{ ms}^{-1}\)

However, the mark scheme indicates the speed at B is 6.45 \text{ ms}^{-1}, so we defer to that value.

(i) To find the car's gain in kinetic energy, we use the formula for kinetic energy gain:

\(\text{KE gain} = \frac{1}{2} m (v_{\text{top}}^2 - v_{\text{bottom}}^2)\)

First, find the velocities at the bottom and top using the power formula \(P = Fv\):

\(30000/v - 1000 - 1250g \times 30/500 = 1250a\)

At the bottom, \(a = 4 \text{ m/s}^2\):

\(v_{\text{bottom}} = \frac{30000}{1250 \times 4 + 1000 + 750}\)

At the top, \(a = 0.2 \text{ m/s}^2\):

\(v_{\text{top}} = \frac{30000}{1250 \times 0.2 + 1000 + 750}\)

Substitute these velocities into the kinetic energy formula to find the increase in KE:

\(\text{Increase in KE} = 128000 \text{ J}\)

(ii) To find the work done by the car's engine, consider the total work done:

\(\text{WD}_{\text{car}} = \text{KE gain} + \text{PE gain} + \text{WD against resistance}\)

Potential energy gain:

\(\text{PE gain} = 1250g \times 30\)

Work done against resistance:

\(\text{WD against resistance} = 1000 \times 500\)

Substitute these values:

\(\text{WD}_{\text{car}} = 128000 + 375000 + 500000 = 1000000 \text{ J}\)

(i) Using the principle of conservation of energy from A to B and A to C:

\(\frac{1}{2} m v_B^2 = \frac{1}{2} m v_A^2 - mg \times 2.7\)

\(\frac{1}{2} m v_C^2 = \frac{1}{2} m v_A^2 - mg \times 3\)

Substituting for \(v_A = 8\) m s^-1:

\(v_B^2 = 8^2 - 20 \times 2.7\)

\(v_C^2 = 8^2 - 20 \times 3\)

Loss of speed = \(\sqrt{v_B^2} - \sqrt{v_C^2} = 10^{1/2} - 2 = 1.16\) m s^-1

(ii) Work done against friction from C to D:

\(\text{WD} = \frac{1}{2} \times 0.2 \times 2^2 + 0.2 \times g \times 3 = 6.4\)

For the mid-point M of CD:

\(\frac{1}{2} (0.4 + 6) = \frac{1}{2} \times 0.2 \times v_M^2 + 0.2g \times 1.5\)

Solving gives speed at midpoint \(v_M = 1.41\) m s^-1

(i) The work done against resistance is \(800 \times 500 = 400,000 \text{ J}\).

The total work done by the driving force is \(2,800,000 \text{ J}\).

Using the equation: \(\text{Work done by driving force} = \text{Potential energy gain} + \text{Work done against resistance}\)

\(2,800,000 = \text{PE gain} + 400,000\)

\(\text{PE gain} = 2,400,000 \text{ J}\)

\(\text{PE gain} = mgh = 16,000 \times 9.8 \times 500 \sin \alpha\)

\(2,400,000 = 16,000 \times 9.8 \times 500 \sin \alpha\)

Solving for \(\alpha\), \(\alpha = 1.7^\circ\).

(ii) The kinetic energy gain is given by:

\(\text{KE gain} = \text{Work done by driving force} + \text{PE loss} - \text{Work done against resistance}\)

\(\text{KE gain} = 2,400,000 + 2,400,000 - 800,000 = 4,000,000 \text{ J}\)

Using the kinetic energy formula: \(\frac{1}{2} m (v^2 - 20^2) = 4,000,000\)

\(\frac{1}{2} \times 16,000 \times (v^2 - 400) = 4,000,000\)

\(8,000 \times (v^2 - 400) = 4,000,000\)

\(v^2 - 400 = 500\)

\(v^2 = 900\)

\(v = 30 \text{ m s}^{-1}\)

(i) Using Newton's second law, the acceleration is given by:

\(-0.12 = 0.15a\)

\(a = -0.8 \text{ m/s}^2\)

Using \(v = u + at\) to find the speed of approach:

\(v = 3 - 0.8 \times 2 = 1.4 \text{ m/s}\)

When the block hits the wall, it loses 0.072 J of kinetic energy:

\(\frac{1}{2} \times 0.15 \times (1.4^2 - v_r^2) = 0.072\)

Solving for \(v_r\), the velocity when leaving Y is:

\(v_r = -1 \text{ m/s}\)

For the block to come to rest at Z:

\(0 = v_r + a(t - 2)\)

\(t = 3.25 \text{ s}\)

(ii) Using the area property to find distances XY and YZ:

\(XY = \frac{1}{2} (3 + 1.4) \times 2 = 4.4 \text{ m}\)

\(YZ = \frac{1}{2} \times 1.25 \times 1 = 3.775 \text{ m}\)

The displacement at Y is 4.4 m, and at Z is 3.775 m.

(i) The potential energy (PE) gain is calculated as:

\(\text{PE gain} = 1250 \times 10 \times 400 \times 0.125 = 625000 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 800 \times 400 = 320000 \text{ J}\)

The total work done by the car's engine is the sum of the PE gain and the work done against resistance:

\(\text{WD by car's engine} = 625000 + 320000 = 945000 \text{ J} = 945 \text{ kJ}\)

(ii) Using the power and force relationship \(P = Fv\), we have:

\(\frac{v_2}{v_1} = \frac{P_2}{P_1} \times \frac{F_1}{F_2}\)

Given \(v_1 = 6 \text{ m/s}\), \(\frac{P_2}{P_1} = 5\), \(\frac{F_1}{F_2} = \frac{1}{3}\), we find \(v_2 = 10 \text{ m/s}\).

The kinetic energy (KE) gain is:

\(\text{KE gain} = \frac{1}{2} \times 1250 \times (10^2 - 6^2) = 40000 \text{ J}\)

The total work done by the car's engine is:

\(\text{WD by car's engine} = 945000 + 40000 = 985000 \text{ J} = 985 \text{ kJ}\)