(a) For the engine, apply Newton's second law:

\(125000 - 120000g \times 0.02 - 22000 - T = 120000a\)

\(125000 - 24000 - 22000 - T = 120000a\)

\(79000 - T = 120000a\)

For the coach:

\(T - 60000g \times 0.02 - 13000 = 60000a\)

\(T - 12000 - 13000 = 60000a\)

\(T - 25000 = 60000a\)

Combine the equations:

\(125000 - 120000g \times 0.02 - 22000 - 13000 = (120000 + 60000)a\)

\(125000 - 24000 - 22000 - 13000 = 180000a\)

\(66000 = 180000a\)

\(a = 0.3 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T - 25000 = 60000 \times 0.3\)

\(T = 43000 \text{ N}\)

(b) Given power \(P = 4500000 \text{ W}\) and speed \(v = 30 \text{ m/s}\), find the driving force \(DF\):

\(DF = \frac{P}{v} = \frac{4500000}{30} = 150000 \text{ N}\)

Resolve forces for the system:

\(150000 - 120000g \sin \beta - 60000g \sin \beta - 22000 - 13000 = 0\)

\(150000 - 180000g \sin \beta - 35000 = 0\)

\(115000 = 180000g \sin \beta\)

\(\sin \beta = \frac{115000}{180000g}\)

\(\beta = 3.7^\circ\)

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

\(\text{Work Done} = \text{Force} \times \text{Distance}\)

The distance covered in 8 seconds at 36 m/s is:

\(36 \times 8 = 288 \text{ m}\)

Thus, the work done is:

\(1250 \times 288 = 360000 \text{ J} = 360 \text{ kJ}\)

(a)(ii) The power developed by the engine is:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{360000}{8} = 45000 \text{ W} = 45 \text{ kW}\)

(a)(iii) With the power increased by 12 kW, the new power is:

\(57000 \text{ W}\)

Using the formula for power:

\(P = DF \times v\)

We find the driving force (DF):

\(DF = \frac{57000}{36} = 1583.3 \text{ N}\)

Using Newton's second law:

\(1583.3 - 1250 = 1400a\)

Solving for acceleration:

\(a = \frac{333.3}{1400} = 0.238 \text{ m s}^{-2}\)

(b) For the inclined section, using the power equation:

\(64000 = 1250 \times 32 + 1400g \sin \theta \times 32\)

Solving for \(\theta\):

\(64000 = 40000 + 44800 \sin \theta\)

\(24000 = 44800 \sin \theta\)

\(\sin \theta = \frac{24000}{44800} = 0.5357\)

\(\theta = \arcsin(0.5357) \approx 3.1^\circ\)

(a) The power exerted by the cyclist is given by \(P = Fv\), where \(F\) is the forward force and \(v\) is the velocity. Thus, \(F = \frac{150}{4} = 37.5 \text{ N}\).

Using Newton's second law, \(F - 20 = m \times 0.25\).

Substitute \(F = 37.5\):

\(37.5 - 20 = m \times 0.25\)

\(17.5 = m \times 0.25\)

\(m = \frac{17.5}{0.25} = 70 \text{ kg}\).

(b) Resolving forces up the plane, the equation is:

\(\frac{150}{3} - 20 - 70g \sin \theta = 0\).

Solving for \(\theta\):

\(50 - 20 = 70g \sin \theta\)

\(30 = 70g \sin \theta\)

\(\sin \theta = \frac{30}{70g}\)

\(\theta = \sin^{-1} \left( \frac{3}{m} \right)\)

\(\theta = 2.5^\circ\) to 1 d.p.

(a) The driving force \(DF\) is given by \(DF = \frac{22500}{v}\). Applying Newton's second law for constant speed (\(a = 0\)), we have:

\(DF - 1400g \times 0.1 - 600 = 0\)

Solving for \(v\), we find \(v = 11.25 \text{ m s}^{-1}\).

(b) For the section inclined at 2°, using Newton's second law:

\(DF - 1400g \sin 2^{\circ} - 600 = 1400a\)

Substitute \(DF = \frac{22500}{11.25}\) and solve for \(a\):

\(\frac{22500}{11.25} - 1400g \sin 2^{\circ} - 600 = 1400a\)

\(a = 0.651 \text{ m s}^{-2}\) (3sf)

(a)(i) The total resistance force is the sum of the resistances on the car and caravan:

\(F = 400 + 250 = 650 \text{ N}\)

The power \(P\) is given by \(P = Fv\), where \(v = 25 \text{ m s}^{-1}\). Thus,

\(P = 650 \times 25 = 16250 \text{ W} = 16.25 \text{ kW}\)

(a)(ii) The engine's power is increased to 39 kW, so the driving force \(DF\) is:

\(DF = \frac{39000}{25} = 1560 \text{ N}\)

Using Newton's second law for the system:

\(1560 - 650 = 2400a\)

Solving for \(a\):

\(a = \frac{1560 - 650}{2400} = 0.379 \text{ m s}^{-2}\)

For the tension \(T\) in the tow-bar:

\(T - 250 = 800a\)

Substituting \(a\):

\(T = 250 + 800 \times 0.379 = 553 \text{ N}\)

(b) The car and caravan travel up a hill inclined at \(\sin^{-1}(0.05)\). The driving force is:

\(DF = 650 + 2400 \times 10 \times 0.05\)

Using \(P = Fv\):

\(32500 = (650 + 2400 \times 0.05)v\)

Solving for \(v\):

\(v = \frac{32500}{650 + 120} = 17.6 \text{ m s}^{-1}\)

(a) Apply Newton's second law: The net force on the car is given by the equation:

\(DF - 650 = 1800 \times 0.5\)

Solving for DF gives:

\(DF = 1800 \times 0.5 + 650 = 1550\)

The power P is given by:

\(\frac{P}{20} - 650 = 1800 \times 0.5\)

Solving for P gives:

\(P = 1550 \times 20 = 31,000 ext{ W or } 31 ext{ kW}\)

(b) To find the steady speed, use the power equation:

\(\frac{31,000}{v} - 650 = 0\)

Solving for v gives:

\(v = \frac{31,000}{650} = 47.7 ext{ m s}^{-1}\)

(a) The power \(P\) is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

Here, \(F = 350 \text{ N}\) and \(v = 20 \text{ m s}^{-1}\).

Thus, \(P = 350 \times 20 = 7000 \text{ W} = 7 \text{ kW}\).

(b) The power \(P\) is 15 kW, which is \(15000 \text{ W}\).

Using \(P = Fv\), we have \(15000 = DF \times 20\), where \(DF\) is the driving force.

Solving for \(DF\), we get \(DF = 750 \text{ N}\).

Using Newton's second law, \(DF - 350 = 1400a\).

Substituting \(DF = 750\), we have \(750 - 350 = 1400a\).

Thus, \(400 = 1400a\).

Solving for \(a\), we get \(a = \frac{400}{1400} = \frac{2}{7} \approx 0.286 \text{ m s}^{-2}\).

(a) Using Newton's second law, the net force is given by:

\(F - 900 = 4000 imes 0.5\)

Solving for the driving force, F:

\(F = 900 + 4000 imes 0.5 = 2900 ext{ N}\)

(b) The power required to maintain a constant speed is given by:

\(P = F imes v\)

Here, the force F is the resistance, 900 N, and the speed v is 25 m/s:

\(P = 900 imes 25 = 22500 ext{ W} ext{ or } 22.5 ext{ kW}\)

(a)(i) The power developed by the engine is given by the formula:

\(P = F \times v\)

where \(F = 750 \text{ N}\) and \(v = 32 \text{ m/s}\).

\(P = 750 \times 32 = 24000 \text{ W} = 24 \text{ kW}\).

(a)(ii) The initial power is 24 kW, and it is decreased by 8 kW, so the new power is:

\(16000 \text{ W} = DF \times 32\)

\(DF = 500 \text{ N}\)

The net force is:

\(500 - 750 = 1250 \times a\)

\(a = -0.2 \text{ m/s}^2\)

(b) The resistance to motion is given by:

\(DF = 1000 + 8v + 1250 \times 10 \times 0.096\)

\(DF = 2200 + 8v\)

The power equation is:

\(60000 = (2200 + 8v) \times v\)

\(8v^2 + 2200v - 60000 = 0\)

Solving this quadratic equation gives:

\(v = 25 \text{ m/s}\)

(a) To find the tension \(T\) in the tow-bar, consider the forces acting on the trailer. The trailer has a mass of 400 kg and is accelerating at 1.5 \(\text{ms}^{-2}\). The net force on the trailer is given by Newton's second law:

\(T - 100 = 400 \times 1.5\)

Solving for \(T\):

\(T = 400 \times 1.5 + 100 = 700 \text{ N}\)

(b) To find the power of the engine, first calculate the total force \(F\) exerted by the engine. The car and trailer together have a total mass of 2200 kg. Using Newton's second law for the entire system:

\(F - 250 - 100 = 2200 \times 1.5\)

\(F = 2200 \times 1.5 + 250 + 100 = 3650 \text{ N}\)

The power \(P\) is given by the product of the force and the velocity:

\(P = Fv = 3650 \times 20 = 73,000 \text{ W}\)

Therefore, the power of the engine is 73,000 W or 73 kW.

(a) The power of the lorry's engine can be calculated using the formula:

\(P = \frac{\text{Work Done}}{\text{Time}}\)

Given that the work done is 750000 J and the time is 10 s, we have:

\(P = \frac{750000}{10} = 75000 \text{ W}\)

Therefore, the power is 75 kW.

(b) To find the acceleration, we first find the driving force (DF) using the power formula:

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

\(75000 = \text{DF} \times 25\)

\(\text{DF} = \frac{75000}{25} = 3000 \text{ N}\)

Using Newton's second law, \(F = ma\), where the net force \(F\) is the driving force minus the resistance force:

\(3000 - 2400 = 16000a\)

\(600 = 16000a\)

\(a = \frac{600}{16000} = 0.0375 \text{ m s}^{-2}\)

(a)(i) The power developed by the engine is given by the formula:

\(P = F \times v\)

where \(F = 650 \text{ N}\) and \(v = 30 \text{ m/s}\).

\(P = 650 \times 30 = 19500 \text{ W} = 19.5 \text{ kW}\).

(a)(ii) The increase in power is 9 kW, so the new power is:

\(19500 + 9000 = DF \times 30\)

\(DF = 1300a + 650\)

Solving for \(a\):

\(9000 = DF \times 30\)

\(DF = 1300a\)

\(a = \frac{3}{13} = 0.231 \text{ m/s}^2\)

(b) Using the power equation:

\(DF \times v = 11500\)

Using Newton's second law:

\(11500/v + 1300 \times g \times 0.08 - (1000 + 20v) = 0\)

Solving for \(v\):

\(v = 25 \text{ m/s}\)

(i) The power output is given by the formula:

\(P = F \times v\)

where \(P = 240\) W and \(v = 6\) m/s. Thus, the driving force \(F\) is:

\(F = \frac{240}{6} = 40\) N

Using Newton's Second Law:

\(F - R = ma\)

where \(m = 80\) kg and \(a = 0.3\) m/s². Therefore:

\(40 - R = 80 \times 0.3\)

\(R = 40 - 24 = 16\) N

(ii) For steady speed, the driving force equals the resistance:

\(\frac{240}{v} = 16\)

Solving for \(v\):

\(v = \frac{240}{16} = 15\) m/s

(iii) On an incline, using Newton's Second Law:

\(\frac{240}{4} - 16 - 80g \sin 3^{\circ} = 80a\)

\(60 - 16 - 80 \times 9.8 \times \sin 3^{\circ} = 80a\)

\(60 - 16 - 41.04 = 80a\)

\(2.96 = 80a\)

\(a = \frac{2.96}{80} = 0.0266\) m/s²

The power \(P\) of the crane is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

The force \(F\) required to lift the load is equal to the weight of the load, which is \(F = mg\), where \(m = 1250\) kg is the mass and \(g = 9.8\) m s^-2 is the acceleration due to gravity.

Thus, \(F = 1250 \times 9.8\).

Given that the power \(P = 20,000\) W (since 20 kW = 20,000 W), we have:

\(20,000 = V \times 1250 \times 9.8\).

Solving for \(V\):

\(V = \frac{20,000}{1250 \times 9.8}\).

\(V = 1.6\) m s^-1.

(i) Using Newton's Second Law, the equation of motion up the hill is:

\(DF - 1550 - 1400g \sin 4^{\circ} = 1400 \times 0.4\)

Solving for the driving force \(DF\):

\(DF = 3086.59 \ldots \text{ N}\)

Using the power formula \(P = Fv\), where \(P = 30000 \text{ W}\):

\(30000 = (1400 \times 0.4 + 1550 + 1400g \sin 4^{\circ})v\)

Solving for \(v\):

\(v = 9.72 \text{ m s}^{-1}\)

(ii) For constant speed, the net force is zero:

\(DF - 1550 - 1400g \sin 4^{\circ} = 0\)

Solving for \(DF\):

\(DF = 2526.59 \ldots \text{ N}\)

Using the power formula \(P = Fv\) with \(v = 40 \text{ m s}^{-1}\):

\(P_{\text{max}} = (1550 + 1400g \sin 4^{\circ}) \times 40\)

\(P = 101063.6 \ldots \text{ W}\)

Thus, the maximum possible power is 101 kW.

First, calculate the driving force (DF) for each case using the formula \(DF = \frac{P}{v}\), where \(P\) is the power and \(v\) is the speed.

Case 1: Horizontal road

\(DF = \frac{36000}{18} = 2000 \text{ N}\)

The resistance force is \(Av + B\), so:

\(18A + B = 2000\)

Case 2: Inclined road

\(DF = \frac{21000}{12} = 1750 \text{ N}\)

The resistance force is \(Av + B\) plus the component of weight along the incline:

\(12A + B + 1000g \times 0.05 = 1750\)

\(12A + B + 500 = 1750\)

\(12A + B = 1250\)

Now solve the simultaneous equations:

1. \(18A + B = 2000\)

2. \(12A + B = 1250\)

Subtract equation 2 from equation 1:

\((18A + B) - (12A + B) = 2000 - 1250\)

\(6A = 750\)

\(A = 125\)

Substitute \(A = 125\) into equation 1:

\(18(125) + B = 2000\)

\(2250 + B = 2000\)

\(B = -250\)

(i) The driving force (DF) required to move the lorry up the hill is given by the sum of the resistance to motion and the component of the weight of the lorry acting down the slope. The component of the weight down the slope is \(12,000 \times g \times 0.08\), where \(g\) is the acceleration due to gravity (approximately 9.8 m s^-2). Therefore,

\(\text{DF} = 1500 + 12,000 \times 9.8 \times 0.08 = 11,100 \text{ N}\)

The power \(P\) of the engine is given by \(P = \text{DF} \times v\), where \(v = 5 \text{ m s}^{-1}\). Thus,

\(P = 11,100 \times 5 = 55,500 \text{ W} = 55.5 \text{ kW}\)

(ii) Given that the resistance to motion is \(kv^2\) and at \(v = 5 \text{ m s}^{-1}\), the resistance is 1500 N, we have:

\(k \times 5^2 = 1500\)

\(25k = 1500\)

\(k = 60\)

(iii) On a straight level road, the driving force is equal to the resistance, which is \(60v^2\). The power is still 55.5 kW, so:

\(55,500 = 60v^2 \times v = 60v^3\)

\(v^3 = \frac{55,500}{60}\)

\(v^3 = 925\)

\(v = 9.74 \text{ m s}^{-1}\)

(i) The driving force \(F\) is given by \(F = \frac{P}{v}\), where \(P = 6000 \text{ W}\) and \(v = 20 \text{ m s}^{-1}\). Thus, \(F = \frac{6000}{20} = 300 \text{ N}\).

The net force on the system is zero since the speed is constant: \(300 - R - 80 = 0\).

Solving for \(R\), we get \(R = 220 \text{ N}\).

(ii) The new driving force \(DF\) is \(\frac{12500}{25} = 500 \text{ N}\).

For the car: \(DF - T - R = 1500a\).

For the trailer: \(T - 80 = 300a\).

For the system: \(DF - 80 - R = 1800a\).

Using \(DF = 500 \text{ N}\) and \(R = 220 \text{ N}\), solve the equations:

1. \(500 - T - 220 = 1500a\)

2. \(T - 80 = 300a\)

Solving these simultaneously gives \(a = 0.111 \text{ m s}^{-2}\) and \(T = 113 \text{ N}\).

(i) The driving force is given by the power equation: \(Fv = 36000\). Therefore, \(F = \frac{36000}{20} = 1800 \text{ N}\). Using Newton's Second Law: \(1800 - R = 3200 \times 0.2\). Solving for \(R\), we get \(R = 1160 \text{ N}\).

(ii) The driving force required to maintain the speed on the incline is \(F = 3200g\sin(1.5^{\circ}) + 1160\). The power required is \(P = (3200g\sin(1.5^{\circ}) + 1160) \times 30\). Calculating gives \(P = 59900 \text{ W}\).

(iii) Using Newton's Second Law: \(- (3200g\sin(1.5^{\circ}) + 1160) = 3200a\). Solving for \(a\), we find \(a = -0.62426 \text{ m/s}^2\). Using the equation \(0^2 = 30^2 + 2as\), solve for \(s\) to get \(s = 721 \text{ m}\).

(i) The power of the car's engine is given by the formula:

\(\text{Power} = \text{Force} \times \text{Velocity}\)

Substituting the given values:

\(\text{Power} = 350 \times 15 = 5250 \text{ W}\)

(ii) When the car descends the hill, the forces acting along the slope are:

\(\text{Driving Force (DF)} + 1200g \sin 1^{\circ} - 350 = 1200 \times 0.12\)

Solving for DF:

\(\text{DF} = 1200 \times 0.12 - 1200g \sin 1^{\circ} + 350\)

The power is then:

\(P = \text{DF} \times 15 = 4270 \text{ W}\)

(iii) Using Newton's second law down the slope:

\(1200g \sin 1^{\circ} - 350 = 1200a\)

Using the constant acceleration formula:

\(18^2 = 20^2 + 2as\)

Solving for \(s\):

\(s = 324 \text{ m}\)

Alternatively, using energy methods:

Potential energy loss: \(1200g \times s \times \sin 1^{\circ}\)

Kinetic energy loss: \(\frac{1}{2} \times 1200 \times (20^2 - 18^2)\)

Equating energy losses to work done against resistance:

\(1200g \times s \times \sin 1^{\circ} + \frac{1}{2} \times 1200 \times (20^2 - 18^2) = 350s\)

Solving gives \(s = 324 \text{ m}\).

(i) The power supplied by the engines is 4080 kW, which is 4080000 W. The speed of the train is 85 m s^-1. Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, we have:

\(F = \frac{P}{v} = \frac{4080000}{85} = 48000 \text{ N}\)

Thus, the resistance force is 48,000 N.

(ii) On the incline, the driving force \(DF\) is given by:

\(DF = \frac{P}{v}\)

Using Newton's second law, the net force is zero for constant speed, so:

\(DF - 48000 - 490000 \times g \times \frac{1}{200} = 0\)

Solving for \(DF\):

\(DF = 48000 + 490000 \times 9.8 \times \frac{1}{200}\)

\(DF = 48000 + 24010 = 72010 \text{ N}\)

The power required is:

\(P = DF \times v = 72010 \times 85 = 6120850 \text{ W} = 6.16 \text{ MW}\)

(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the resistive force. At the greatest speed, \(v = 56 \text{ m s}^{-1}\), the resistive force \(F = 40 \times 56\). Therefore, \(P = 40 \times 56 \times 56 = 125440 \text{ W} = 125 \text{ kW}\).

(ii) The driving force \(F_d\) is \(\frac{125440}{32}\). Using Newton's second law, \(F_d - 40 \times 32 = 1400a\). Solving for \(a\), we get \(a = 1.89 \text{ m s}^{-2}\).

(iii) For the car on an incline, the equation is \(\frac{60000}{50} + 1400g \sin \theta - 40 \times 50 = 0\). Solving for \(\sin \theta\), we have \(\sin \theta = \frac{800}{14000}\). Thus, \(\theta = 3.3^\circ\).

(a) To find the power of the lorry's engine, use the formula for power: \(P = Fv\). At point A, \(P = 20F\) and at point B, \(P = 25F\). Given the resistance force is 6000 N, the net force at A is \(F - 6000\) and at B is \(F - 6000\).

Using Newton's second law, at A: \(\frac{P}{20} - 6000 = 15000a\) and at B: \(\frac{P}{25} - 6000 = 15000 \times 0.5a\).

Solving these equations simultaneously, we find \(P = 200,000 \text{ W} = 200 \text{ kW}\) and \(a = \frac{4}{15} \text{ m/s}^2\).

(b) For the lorry ascending the hill, the equation becomes \(\frac{200,000}{v} - 6000 - 15000g \sin(1^{\circ}) = 0\). Solving for \(v\), we find the steady speed \(v = 23.2 \text{ m/s}\).

To find the power of the engine, we first need to determine the driving force (DF) using the equation for power:

\(DF = \frac{P}{v}\)

where \(P\) is the power and \(v = 15 \text{ m/s}\) is the speed of the train.

Using Newton's second law, the equation for the forces acting on the train is:

\(DF - 240,000g \sin 4^{\circ} - 18,000 = 240,000 \times (-0.2)\)

Solving for \(DF\):

\(DF = 240,000 \times 9.8 \times \sin 4^{\circ} + 18,000 - 240,000 \times 0.2\)

Substitute \(DF\) back into the power equation:

\(P = DF \times 15\)

Calculate \(P\) to find:

\(P = 2,060,000 \text{ W}\)

(i) The driving force when the car is at its greatest speed is given by the resistive force: \(35 \times 60\). The power is then \(35 \times 60^2 = 126000 \text{ W}\).

(ii) The driving force \(\text{DF}\) is \(\frac{126000}{30}\). Using Newton's second law: \(\text{DF} - 35 \times 30 = 1200a\). Solving for \(a\), we have \(a = \frac{3150}{1200} = \frac{21}{8} = 2.625 \text{ m s}^{-2}\).

(iii) The driving force \(\text{DF} = \frac{126000}{v}\). The equation for forces is \(\frac{126000}{v} = 35v + 1200g \times \frac{7}{48}\). Simplifying gives \(35v^2 + 1750v - 126000 = 0\) or \(v^2 + 50v - 3600 = 0\). Solving this quadratic equation gives \(v = 40 \text{ m s}^{-1}\).

(i) When the lorry is going up the hill at a constant speed, the force required is given by:

\(F = 1480 + 7850g \sin 3^{\circ}\)

Calculating the force:

\(F = 1480 + 7850 \times 9.8 \times \sin 3^{\circ} = 5588 \text{ N}\)

The power is given by:

\(P = Fv = 5588 \times 10 = 55,880 \text{ W}\)

Rounding to three significant figures, the power is 55,900 W.

(ii) When the lorry is going down the hill with acceleration, use Newton's Second Law:

\(F + 7850g \sin 3^{\circ} - 1480 = 7850 \times 0.8\)

Solving for the force:

\(F = 3652 \text{ N}\)

The power is given by:

\(P = Fv = 3652 \times 15 = 54,780 \text{ W}\)

Rounding to three significant figures, the power is 54,800 W.

(i) To find the power output of the cyclist, we first calculate the resistance force: \(F_{\text{resistance}} = \frac{600}{25} = 24 \text{ N}\).

The weight component along the incline is \(80g \sin \alpha = 80 \times 9.8 \times 0.04 = 32 \text{ N}\).

The total force exerted by the cyclist is \(24 + 32 = 56 \text{ N}\).

Power is given by \(P = Fv\), where \(F\) is the total force and \(v\) is the velocity. Thus, \(P = 56 \times 4 = 224 \text{ W}\).

(ii) The work done by the cyclist in 10 seconds is \(224 \times 10 = 2240 \text{ J}\).

The initial kinetic energy (KE) at the top of the hill is \(\frac{1}{2} \times 80 \times 4^2 = 640 \text{ J}\).

Using the work-energy principle: \(\frac{1}{2} \times 80 \times v^2 = 640 + 2240 - 1200\).

Solving for \(v\), we get \(v^2 = 42\), so \(v = \sqrt{42} \approx 6.48 \text{ m s}^{-1}\).

(i) The power output of the tractor’s engine is given by the formula:

\(P = F \times v\)

where \(F = 1150 \text{ N}\) and \(v = 12 \text{ m s}^{-1}\).

Thus, \(P = 1150 \times 12 = 13,800 \text{ W}\).

(ii) The driving force when the power is increased to 25 kW is:

\(F = \frac{25000}{12}\)

Using Newton’s second law up the slope:

\(\frac{25000}{12} - 1150 - 3700g \sin 4^{\circ} = 3700a\)

Solving for \(a\):

\(a = -0.445 \text{ m s}^{-2}\)

(iii) For constant speed \(v\) on the hill:

\(\frac{25000}{v} - 1150 - 3700g \sin 4^{\circ} = 0\)

Solving for \(v\):

\(v = 6.70 \text{ m s}^{-1}\)

(i)(a) Using the formula for power, \(P = Fv\), where \(P = 16000\) W and \(v = 40\) m/s, we have:

\(16000 = F \times 40\)

Solving for \(F\), we get \(F = 400\) N.

(i)(b) With the increased power, \(P = 22500\) W and \(v = 45\) m/s, we find the force:

\(22500 = F \times 45\)

\(F = 500\) N.

Using Newton's Second Law, \(F - ext{resistance} = ma\):

\(500 - 400 = 1200a\)

\(a = \frac{1}{12} = 0.0833 \text{ m/s}^2\)

(ii) Given \(16000 = (590 + 2v)v\), solve for \(v\):

\(16000 = 590v + 2v^2\)

\(2v^2 + 590v - 16000 = 0\)

Solving this quadratic equation gives \(v = 25 \text{ m/s}\).

(i)(a) The power developed by the engine is given by the formula:

\(P = Fv\)

where \(F = 850 \text{ N}\) and \(v = 42 \text{ m s}^{-1}\).

Thus, \(P = 850 \times 42 = 35700 \text{ W} = 35.7 \text{ kW}\).

(i)(b) The new power is \(35700 + 6000 = 41700 \text{ W}\).

The driving force \(DF\) is given by:

\(DF = \frac{41700}{42} = 993 \text{ N}\).

Using Newton's second law:

\(993 - 850 = 1200a\)

\(143 = 1200a\)

\(a = \frac{143}{1200} = 0.119 \text{ m s}^{-2}\).

(ii) The driving force \(DF\) is given by:

\(DF = \frac{80000}{24} = 3333.33 \text{ N}\).

Using Newton's second law along the incline:

\(DF - 850 - 1200g \sin \theta = 0\)

\(3333.33 - 850 = 1200 \times 9.8 \times \sin \theta\)

\(2483.33 = 11760 \sin \theta\)

\(\sin \theta = \frac{2483.33}{11760}\)

\(\theta = \sin^{-1} \left( \frac{2483.33}{11760} \right) \approx 11.9^\circ\).

(i) Using the formula for power, \(P = Fv\), where \(P = 36000\) W and \(F = 800\) N, we have:

\(36000 = 800v\)

\(v = \frac{36000}{800} = 45 \text{ m/s}\)

The distance \(AB\) is given by \(v \times t = 45 \times 120 = 5400 \text{ m}\).

(ii) Using Newton's second law, \(-800 = 900a\), we find \(a = -\frac{8}{9} \text{ m/s}^2\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 45 \text{ m/s}\), \(s = 450 \text{ m}\):

\(v^2 = 45^2 - \frac{16}{9} \times 450\)

\(v = 35 \text{ m/s}\)

(iii) The distance \(CD = 6637.5 - 5400 - 450 = 787.5 \text{ m}\).

Using \(v^2 = u^2 + 2as\), where \(u = 35 \text{ m/s}\), \(v = 0\):

\(0 = 35^2 - 2d \times 787.5\)

\(d = \frac{7}{9} = 0.778 \text{ m/s}^2\)

Using \(F = ma\), \(R = 900 \times \frac{7}{9} = 700 \text{ N}\).

(i) The driving force is given by the power divided by the speed: \(\frac{160}{5} = 32 \text{ N}\). Using Newton's Second Law, \(32 - 20 = m \times 0.15\). Solving for \(m\), we get \(m = \frac{12}{0.15} = 80 \text{ kg}\).

(ii) For the hill, resolve forces: \(\frac{300}{v} - 20 - 80g \sin 2^\circ = 0\). Solving for \(v\), we find \(v = 6.26 \text{ m s}^{-1}\).

(iii) The driving force at 90% of the speed is \(\frac{300}{0.9 \times 6.26} = 53.2 \text{ N}\). Using Newton's Second Law: \(53.2 - 20 - 80g \sin 2^\circ = 80a\). Solving for \(a\), we get \(a = 0.0666 \text{ m s}^{-2}\).

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

\(\text{PE gain} = 50g \times 3.5 = 1750 \text{ J}\)

The work done (WD) by the crane is the sum of the PE gain and the work done against resistance:

\(\text{WD} = 50g \times 3.5 + 25 \times 3.5 = 1837.5 \text{ J}\)

Thus, the work done by the crane is 1837.5 J or 1840 J.

(ii) The power (P) of the crane is calculated using the formula:

\(P = \frac{\text{WD}}{t} = \frac{1837.5}{2} = 918.75 \text{ W}\)

Rounding gives the power as 919 W.

(a) Since the locomotive and truck are moving at a constant speed, the net force is zero. Therefore, the tension in the coupling is 0 N.

(b) The power produced by the locomotive's engine is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. The resistance force is 0.2 N, and the velocity is 2 m/s. Thus, \(P = 0.2 \times 2 = 0.4 \text{ W}\).

(c) When the power is changed to 1.2 W, the driving force \(DF\) is \(\frac{1.2}{2} = 0.6 \text{ N}\). Using Newton's second law for the locomotive: \(DF - 0.2 - T = 0.8a\). For the truck: \(T = 0.4a\). For the system: \(DF - 0.2 = 1.2a\). Solving these equations gives \(T = \frac{2}{15} \text{ N}\).

(i) The power of the van's engine is given by the formula:

\(P = Fv\)

where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force is \(300 + 100 = 400\) N. Therefore,

\(P = 400 \times 25 = 10,000\) W.

(ii) The tension in the cable is equal to the resistance force on the trailer, which is 100 N.

(iii) When the van is moving up the hill, the new driving force is calculated as:

\(\text{Driving force} = \frac{25,000}{20} = 1,250\) N.

Using Newton's second law for the system of van and trailer:

\(1,250 - 300 - T - 3000 \sin 4^{\circ} = 3000a\)

or for the trailer:

\(T - 100 - 500 \sin 4^{\circ} = 500a\)

Solving these equations, we find:

\(T = 221\) N.

(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the speed. Given \(P = 20000\) W and \(F = 650\) N, we have:

\(20000 = 650v\)

Solving for \(v\), we get:

\(v = \frac{20000}{650} = 30.8 \text{ m s}^{-1}\)

(ii) The force required to move up the hill is the sum of the resistive force and the component of the gravitational force along the hill. This is given by:

\(F = 650 + 1400g \times \frac{1}{7}\)

Where \(g = 9.8 \text{ m s}^{-2}\). The power \(P\) is then:

\(\frac{P}{10} = 650 + 1400 \times 9.8 \times \frac{1}{7}\)

Solving for \(P\), we find:

\(P = 26500 \text{ W}\)

(iii) The power when descending is 80% of the power found in part (ii):

\(P = 0.8 \times 26500 = 21200 \text{ W}\)

Using \(P = Fv\), where \(v = 20 \text{ m s}^{-1}\), we have:

\(21200 = F \times 20\)

\(F = 1060 \text{ N}\)

The net force is given by:

\(1060 - 650 - 1400 \times 9.8 \times \frac{1}{7} = 1400a\)

Solving for \(a\), we get:

\(a = 1.72 \text{ m s}^{-2}\)

(i) (a) The power developed by the engine is given by the formula:

\(P = Fv\)

where \(F = 1550 \text{ N}\) and \(v = 40 \text{ m s}^{-1}\).

\(P = 1550 \times 40 = 62000 \text{ W} = 62 \text{ kW}\).

(b) The new power is \(62 \text{ kW} - 22 \text{ kW} = 40 \text{ kW} = 40000 \text{ W}\).

Let \(DF\) be the new driving force. Then:

\(40000 = DF \times 40\)

\(DF = 1000 \text{ N}\).

Using Newton's second law:

\(DF - 1550 = 1100a\)

\(1000 - 1550 = 1100a\)

\(-550 = 1100a\)

\(a = -0.5 \text{ m s}^{-2}\) or \(d = 0.5 \text{ m s}^{-2}\).

(ii) The driving force \(DF\) is given by:

\(DF = 1100g \sin 8^{\circ} + 1550\)

\(DF = 3081 \text{ N}\).

The power equation is:

\(80000 = 3081v\)

\(v = \frac{80000}{3081} \approx 26 \text{ m s}^{-1}\).

(i) The driving force required to overcome the resistance is 300 N. The power \(P\) is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

Substituting the given values, \(P = 300 \times 40 = 12000 \text{ W} = 12 \text{ kW}\).

(ii) The power at 90% of the value found in part (i) is \(0.9 \times 12000 = 10800 \text{ W}\).

Using the formula \(P = Fv\), we find the driving force \(F\) when the speed is 25 m s^-1: \(F = \frac{10800}{25} = 432 \text{ N}\).

Applying Newton's second law, \(F - 300 = 1000a\), where 300 N is the resistance.

Thus, \(432 - 300 = 1000a\).

Solving for \(a\), we get \(a = \frac{132}{1000} = 0.132 \text{ ms}^{-2}\).

(i) The power \(P\) required to overcome the resistance is given by \(P = F \times v\), where \(F = 1350\) N and \(v = 32\) m/s.

\(P = 1350 \times 32 = 43200 \text{ W} = 43.2 \text{ kW}\).

(ii) When the car travels up the hill, the driving force \(DF\) must overcome both the resistance and the component of the weight down the slope.

The component of the weight down the slope is \(1200g \sin \theta\), where \(g = 9.8 \text{ m/s}^2\).

\(DF - 1350 - 1200 \times 9.8 \times 0.1 = 0\)

\(DF = 1350 + 1200 \times 9.8 \times 0.1 = 2550 \text{ N}\)

The power \(P\) is given by \(P = DF \times v\).

\(76500 = 2550 \times v\)

\(v = \frac{76500}{2550} = 30 \text{ m/s}\)

(i) To show that \(R = \frac{420}{v} + 43.56\):

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

\(a = \frac{5^2 - 3^2}{2 \times 500} = 0.016\)

The force equation is:

\(DF + 90g \times 0.05 - R = 90 \times 0.016\)

Using \(DF = \frac{P}{v}\), we have:

\(R = \frac{420}{v} + 43.56\)

(ii) To find the cyclist’s speed at the mid-point of the hill:

\(v_M^2 = 3^2 + 2 \times 0.016 \times 250\)

The speed at the mid-point is 4.12 m s^-1.

Decrease in \(R\) from top to mid-way:

\(420 \left( \frac{1}{3} - \frac{1}{\sqrt{17}} \right)\)

Decrease in \(R\) from midway to bottom:

\(420 \left( \frac{1}{\sqrt{17}} - \frac{1}{5} \right)\)

The decreases are 38.1 and 17.9 respectively.

(i) Apply Newton's second law to the lorry moving up the hill:

\(F - 24000g \sin 3^{\circ} - 3200 = 24000 \times 0.2\)

Solving for \(F\):

\(F = 24000 \times 9.8 \times \sin 3^{\circ} + 3200 + 24000 \times 0.2 = 20561 \text{ N}\)

The power developed by the engine is given by:

\(P = Fv = 20561 \times 25 = 514025 \text{ W} = 514 \text{ kW}\)

(ii) For steady speed, the net force is zero, so:

\(F = 3200 + 24000g \sin 3^{\circ} = 15761 \text{ N}\)

Using the power formula:

\(P = Fv \Rightarrow v = \frac{P}{F} = \frac{500000}{15761} = 31.7 \text{ m s}^{-1}\)

(i) The work done by the weightlifter is calculated using the formula for gravitational work done:

\(WD = mgh\)

where \(m = 200 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 0.7 \text{ m}\).

Substituting the values, we get:

\(WD = 200 \times 9.8 \times 0.7 = 1400 \text{ J}\)

(ii) The average power developed is calculated using the formula:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}}\)

Substituting the values, we get:

\(\text{Power} = \frac{1400}{1.2} = 1166.67 \text{ W}\)

Rounding to the nearest whole number, the average power is \(1170 \text{ W}\).

Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, and applying Newton's second law \(F = ma\), we have:

At the first point:

\(\frac{P}{4.5} - R = 860 \times 4\)

At the second point:

\(\frac{P}{22.5} - R = 860 \times 0.3\)

Subtract the second equation from the first to eliminate \(R\):

\(\frac{P}{4.5} - \frac{P}{22.5} = 860(4 - 0.3)\)

Solving for \(P\):

\(P = 17900\) W

Substitute \(P\) back into one of the original equations to find \(R\):

\(\frac{17900}{4.5} - R = 860 \times 4\)

\(R = 537.5\) N

The work done by the tension is calculated using the formula:

\(ext{Work Done (WD)} = F imes s\)

where \(F = 500 \text{ N}\) is the tension and \(s\) is the distance traveled.

The distance \(s\) can be found using the speed \(v = 2.75 \text{ m/s}\) and time \(t = 40 \text{ s}\):

\(s = v imes t = 2.75 imes 40 = 110 \text{ m}\)

Thus, the work done is:

\(ext{WD} = 500 imes 110 = 55000 \text{ J}\)

The power applied by the tension is given by:

\(P = \frac{\text{WD}}{t} = \frac{55000}{40} = 1375 \text{ W}\)

Alternatively, power can also be calculated using:

\(P = F imes v = 500 imes 2.75 = 1375 \text{ W}\)

(a) The power of the winch is given as 4.5 kW, which is 4500 W. The work done by the winch is equal to the force times the distance, which is also equal to power times time. Therefore, we have:

\(4500 = \frac{600 \times 15}{t}\)

Solving for \(t\), we get:

\(t = \frac{600 \times 15}{4500} = 2 \text{ s}\)

(b) After the rope breaks, the only force acting on the crate is the resistance of 600 N. Using Newton's second law, \(F = ma\), we have:

\(600 = 200a\)

\(a = 3 \, \text{m/s}^2\)

The initial velocity \(u\) is the velocity at which the crate was moving before the rope broke. From part (a), the velocity \(v\) is:

\(600v = 4500 \Rightarrow v = 7.5 \, \text{m/s}\)

Using the equation \(v = u + at\) with \(v = 0\) (since the crate comes to rest), \(u = 7.5\), and \(a = -3\):

\(0 = 7.5 - 3t\)

Solving for \(t\), we get:

\(t = \frac{7.5}{3} = 2.5 \text{ s}\)

(i) Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. The force \(F\) can be found using Newton's second law \(F = ma\), where \(m = 80\) kg and \(a = 1.2\) m s\(^{-2}\).

Thus, \(F = 80 \times 1.2 = 96\) N.

Substitute into the power formula: \(P = 96 \times 5 = 480\) W.

(ii) The power \(P = 450\) W, speed \(v = 3.6\) m s\(^{-1}\), and \(\sin \alpha = 0.035\).

Using the formula \(\frac{P}{v} - mg \sin \alpha = ma\), where \(m = 80\) kg and \(g = 9.8\) m s\(^{-2}\).

\(\frac{450}{3.6} - 80 \times 9.8 \times 0.035 = 80a\)

\(125 - 27.44 = 80a\)

\(97.56 = 80a\)

\(a = \frac{97.56}{80} = 1.21\) m s\(^{-2}\).

Using the formula for power, \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, we apply Newton's second law for both uphill and downhill motion.

For uphill motion:

\(\frac{P}{3} - R - 84g \times 0.1 = 84 \times 1.25\)

For downhill motion:

\(\frac{P}{10} - R + 84g \times 0.1 = 84 \times 1.25\)

Solving these equations simultaneously:

\(P \left( \frac{1}{3} - \frac{1}{10} \right) - 168 = 0\)

\(P \times \frac{7}{30} = 168\)

\(P = 720\)

Substitute \(P = 720\) into one of the equations to find \(R\):

\(\frac{720}{3} - R - 84 \times 0.1 = 84 \times 1.25\)

\(240 - R - 8.4 = 105\)

\(R = 51\)

(i) The driving force \(DF\) is given by \(DF = \frac{P}{18}\). Using the equation \(DF - R = ma\), where \(R = 800\) N and \(a = 0.5\) m s^-2, we have:

\(\frac{P}{18} - 800 = 1400 \times 0.5\)

\(\frac{P}{18} = 800 + 700\)

\(\frac{P}{18} = 1500\)

\(P = 27000\) W

(ii) At the new speed of 25 m s^-1, the driving force \(DF\) is \(\frac{P}{25}\). Using \(DF - R = ma\), we have:

\(\frac{27000}{25} - 800 = 1400a\)

\(1080 - 800 = 1400a\)

\(280 = 1400a\)

\(a = 0.2\) m s^-2

(i) Let the speed of the train at B be \(v_B\). The power at B is 1.2 times the power at A, and the driving force at B is 0.96 times the driving force at A. Using the formula for power \(P = Fv\), we have:

\(v_B = \frac{1.2 \times 28}{0.96}\)

Calculating this gives \(v_B = 35\) m/s.

(ii) The increase in kinetic energy (KE) from A to B is given by:

\(\text{KE increase} = \frac{1}{2} \times 200,000 \times (35^2 - 28^2)\)

\(= 100,000 \times (35^2 - 28^2)\)

\(= 100,000 \times (1225 - 784)\)

\(= 100,000 \times 441\)

\(= 44.1 \times 10^6 \text{ J}\)

The work done by the engine is the sum of the increase in kinetic energy and the work done against resistance:

\(\text{WD by engine} = 44.1 \times 10^6 + 2.3 \times 10^6\)

\(= 46.4 \times 10^6 \text{ J}\)

Thus, the work done by the engine is 46,400,000 J.

First, calculate the driving force (DF) using the power formula:

\(Power = Force × Velocity\)

\(Given power = 22.5 kW = 22500 W and velocity = 18 m/s,\)

DF = \(\frac{22500}{18} = 1250\) N

Using Newton's second law, the net force is given by:

\(Net Force = Mass × Acceleration\)

Let the resistance force be \(R\). Then,

\(DF - R = 800 × 1.2\)

\(1250 - R = 960\)

\(R = 1250 - 960 = 290 \text{ N}\)

Using Newton's second law, the driving force \(DF\) is given by:

\(DF - 600 - 1250 \times 0.02g = 1250 \times 0.5\)

Solving for \(DF\):

\(DF = 600 + 1250 \times 0.02 \times 9.8 + 1250 \times 0.5\)

\(DF = 600 + 250 + 625 = 1475 \text{ N}\)

The power provided by the engine is \(23 \text{ kW} = 23000 \text{ W}\). Therefore,

\(v = \frac{23000}{1475}\)

\(v \approx 15.6 \text{ m/s}\)

Thus, the speed of the car is 15.6 m/s.

(i) The power \(P\) of the engine is given as \(22.5 \, \text{kW} = 22500 \, \text{W}\). The force \(F\) exerted by the engine is given by \(F = \frac{P}{v}\), where \(v = 18 \, \text{m/s}\). Thus, \(F = \frac{22500}{18} = 1250 \, \text{N}\).

Using Newton's second law, the net force is \(F - R = ma\), where \(m = 600 \, \text{kg}\) and \(a = 1.4 \, \text{m/s}^2\). Therefore, \(1250 - R = 600 \times 1.4\).

Solving for \(R\), we get \(R = 1250 - 840 = 410 \, \text{N}\).

(ii) When the car is moving at a constant speed of \(15 \, \text{m/s}\), the power \(P\) is given by \(P = R \times v\). Thus, \(P = 410 \times 15 = 6150 \, \text{W}\).

The power \(P\) of the train is given by the formula:

\(P = F \times V\)

where \(F\) is the force (resistance) and \(V\) is the speed.

Given:

• \(P = 1330 \text{ kW} = 1330000 \text{ W}\)
• \(F = 28 \text{ kN} = 28000 \text{ N}\)

Substitute these values into the formula:

\(1330000 = 28000 \times V\)

Solve for \(V\):

\(V = \frac{1330000}{28000} = 47.5 \text{ m s}^{-1}\)

(i) Using the formula for power, \(P = Fv\), where \(F\) is the driving force and \(v\) is the velocity, we have:

\(144000 = F \times 16\)

\(F = \frac{144000}{16} = 9000 \text{ N}\)

Using Newton's second law, \(F - 4800 = 12500a\), where \(a\) is the acceleration at A:

\(9000 - 4800 = 12500a\)

\(4200 = 12500a\)

\(a = \frac{4200}{12500} = 0.336 \text{ m s}^{-2}\)

To find the speed at B, use the equation \(v^2 = u^2 + 2as\):

\(v^2 = 16^2 + 2 \times 0.096 \times s\)

Given \(a = 0.096 \text{ m s}^{-2}\) at B, solve for \(v\):

\(v = 24 \text{ m s}^{-1}\)

(ii) Work done by the driving force is \(5800 \times 500\).

Work done against resistance is \(4800 \times 500\).

Loss in kinetic energy is \(\frac{1}{2} \times 12500 \times (24^2 - 16^2)\).

Using energy conservation:

\(5800 \times 500 = 12500gh + \frac{1}{2} \times 12500 \times (24^2 - 16^2) + 4800 \times 500\)

Solving for \(h\):

\(5800 \times 500 = 12500gh + 4800 \times 500 + \frac{1}{2} \times 12500 \times (24^2 - 16^2)\)

\(2900000 = 12500gh + 2400000 + 12500 \times 80\)

\(2900000 = 12500gh + 2400000 + 1000000\)

\(2900000 = 12500gh + 3400000\)

\(12500gh = 2900000 - 3400000\)

\(12500gh = -500000\)

\(h = \frac{-500000}{12500 \times 9.8}\)

\(h = 20 \text{ m}\)

(i) The power exerted by the runner is given by \(P = Fv\), where \(F\) is the force. On horizontal ground, the resistance force \(F = kv\). Therefore, \(P = kv^2\).

Given \(P = 100 \text{ W}\) and \(v = 4 \text{ m s}^{-1}\), we have:

\(100 = k \times 4^2\)

\(100 = 16k\)

\(k = \frac{100}{16} = 6.25\)

(ii) When running uphill, the forces acting on the runner include the resistance \(kv\), gravitational component \(70g \sin \alpha\), and the power equation \(P = Fv\).

The equation becomes:

\(\frac{100}{v} = 70g \times 0.05 + 6.25v\)

\(\frac{100}{v} = 35 + 6.25v\)

Multiplying through by \(v\) gives:

\(100 = 35v + 6.25v^2\)

Rearranging gives the quadratic equation:

\(6.25v^2 + 35v - 100 = 0\)

Solving this quadratic equation using the quadratic formula \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6.25\), \(b = 35\), and \(c = -100\), we find:

\(v = \frac{-35 \pm \sqrt{35^2 - 4 \times 6.25 \times (-100)}}{2 \times 6.25}\)

\(v = \frac{-35 \pm \sqrt{1225 + 2500}}{12.5}\)

\(v = \frac{-35 \pm \sqrt{3725}}{12.5}\)

Calculating gives the maximum speed \(v \approx 2.08 \text{ m s}^{-1}\).

(a)(i) The power of the car’s engine is given by the formula:

\(P = F \times v\)

where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force \(F = 650 + 150 = 800\) N. Therefore,

\(P = 800 \times 24 = 19200\) W or 19.2 kW.

(a)(ii) When the engine’s power is increased to 40 kW, we have:

\(40000 = DF \times 24\)

Using Newton’s Second Law for the system:

\(\frac{40000}{24} - 800 = 2250a\)

Solving for \(a\):

\(a = \frac{40000}{24 \times 2250} - \frac{800}{2250} \approx 0.385\) m s^-2

For the tension \(T\) in the tow-bar:

\(T - 150 = 500a\)

\(T = 500 \times 0.385 + 150 = 343\) N

(b) Resolving forces up the hill, the total resistance force \(DF = 800 + 2250 \times g \times 0.14\).

Using \(P = DF \times v\):

\(31000 = (800 + 2250 \times 9.8 \times 0.14) \times v\)

Solving for \(v\):

\(v = \frac{31000}{800 + 2250 \times 9.8 \times 0.14} \approx 7.85\) m s^-1

To find the values of \(P\) and \(R\), we use the formula for power: \(P = Fv\), where \(F\) is the driving force and \(v\) is the velocity. The driving force \(F\) can be expressed as \(F = ma + R\), where \(m\) is the mass, \(a\) is the acceleration, and \(R\) is the resistance.

For the first condition (speed = 14 m s\(^{-1}\), acceleration = 1.4 m s\(^{-2}\)):

\(1000 \frac{P}{14} - R = 800 \times 1.4\)

For the second condition (speed = 25 m s\(^{-1}\), acceleration = 0.33 m s\(^{-2}\)):

\(1000 \frac{P}{25} - R = 800 \times 0.33\)

Solving these two equations simultaneously:

1. \(1000 \frac{P}{14} - R = 1120\)

2. \(1000 \frac{P}{25} - R = 264\)

Subtract equation 2 from equation 1:

\(1000 \left( \frac{P}{14} - \frac{P}{25} \right) = 1120 - 264\)

\(1000 \left( \frac{25P - 14P}{350} \right) = 856\)

\(1000 \frac{11P}{350} = 856\)

\(11P = 299.6\)

\(P = 27.2 \text{ kW}\)

Substitute \(P = 27.2\) back into one of the original equations to find \(R\):

\(1000 \frac{27.2}{14} - R = 1120\)

\(1942.857 - R = 1120\)

\(R = 1942.857 - 1120\)

\(R = 822.857 \approx 825 \text{ N}\)

(i) The driving force \(F\) provided by the engine is given by \(F = \frac{1000P}{25}\).

Using Newton's second law, the net force is also given by \(F - 600 = 1000 \times 0.2\).

Substituting the expression for \(F\), we have:

\(\frac{1000P}{25} - 600 = 200\)

\(40P - 600 = 200\)

\(40P = 800\)

\(P = 20\)

(ii) For steady speed, the acceleration \(a = 0\). Therefore, the driving force equals the resistance:

\(\frac{1000P}{v_{\text{max}}} = 600\)

Substituting \(P = 20\), we get:

\(\frac{20000}{v_{\text{max}}} = 600\)

\(v_{\text{max}} = \frac{20000}{600}\)

\(v_{\text{max}} = 33.3 \text{ m s}^{-1}\)

(i) The driving force (DF) is given by the formula for power: \(P = Fv\), where \(P = 1500,000\) W and \(v = 37.5\) m/s. Thus, \(DF = \frac{1500,000}{37.5} = 40,000\) N.

Using Newton's second law, \(DF - R = ma\), where \(R = 30,000\) N and \(m = 400,000\) kg.

Substitute the values: \(40,000 - 30,000 = 400,000a\).

Solve for \(a\): \(10,000 = 400,000a\) gives \(a = 0.025\) m/s².

(ii) For steady speed, acceleration \(a = 0\). Thus, \(DF - R = 0\).

\(\frac{1500,000}{v} - 30,000 = 0\).

Solve for \(v\): \(\frac{1500,000}{v} = 30,000\) gives \(v = 50\) m/s.

(i) The driving force (DF) provided by the engine is calculated using power and speed:

\(DF = \frac{17280}{12} = 1440 \text{ N}\).

Using Newton's second law, \(DF - R = ma\), where \(R = 960 \text{ N}\):

\(1440 - 960 = 1200a\).

Solving for \(a\), we get \(a = 0.4 \text{ m/s}^2\).

(ii) For constant speed \(V\), the net force is zero:

\(\frac{17280}{V} - 960 = 0\).

Solving for \(V\), we find \(V = 18 \text{ m/s}\).

(iii) After the engine is switched off, the car decelerates due to resistance:

\(-960 = 1200a\), giving \(a = -0.8 \text{ m/s}^2\).

Using the equations of motion for the segment \(BC\):

\(0 = 18 + (-0.8)t_{BC}\), solving gives \(t_{BC} = 22.5 \text{ s}\).

\(0 = 18^2 + 2(-0.8)s_{BC}\), solving gives \(s_{BC} = 202.5 \text{ m}\).

The distance \(AB\) is \(18 \times (52.5 - 22.5) = 540 \text{ m}\).

Thus, the total distance \(AC = AB + BC = 540 + 202.5 = 742.5 \text{ m}\).

(i) For \(0 < t < 0.5\), the average acceleration \(a_{\text{ave}}\) is calculated as:

\(a_{\text{ave}} = \frac{6 - 4}{0.5 - 0} = \frac{2}{0.5} = 4 \text{ m/s}^2\)

For \(16.3 < t < 24.5\), the average acceleration \(a_{\text{ave}}\) is:

\(a_{\text{ave}} = \frac{21 - 19}{24.5 - 16.3} = \frac{2}{8.2} \approx 0.244 \text{ m/s}^2\)

(ii) Using the accelerations from part (i), assume \(a_1 = 4 \text{ m/s}^2\) at \(v = 5\) and \(a_2 = 0.244 \text{ m/s}^2\) at \(v = 20\).

Using Newton's second law, \(F = ma\), and the given power and resistance:

For \(v = 5\):

\(\frac{P}{5} - R = 1230 \times 4\)

For \(v = 20\):

\(\frac{P}{20} - R = 1230 \times 0.244\)

Solving these equations simultaneously:

\(P = 30800 \text{ W}\)

\(R = 1240 \text{ N}\)

Using Newton's second law, the net force \(F\) acting on the car is given by:

\(F - 700 = 880 \times 0.625\)

Solving for \(F\):

\(F = 880 \times 0.625 + 700 = 1250 \text{ N}\)

The power \(P\) is given by the formula:

\(P = F \times v\)

Substituting the values:

\(P = 1250 \times 16 = 20,000 \text{ W}\)

(i) The power supplied by the engine is 30 kW, which is 30000 W. The work done by the engine is given by:

\(\text{Work Done} = \text{Power} \times \text{Time} = 30000 \times 100 = 3000000 \text{ J}\)

The work done against the resistance is:

\(\text{Work Done} = \text{Resistance} \times \text{Distance} = 750 \times AB\)

Equating the two expressions for work done:

\(750 \times AB = 3000000\)

\(AB = \frac{3000000}{750} = 4000 \text{ m}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\), where \(u = 40 \text{ m/s}\), \(v = 20 \text{ m/s}\), and \(a\) is the acceleration:

\(20^2 = 40^2 + 2(-1.25)BC\)

Solving for \(BC\):

\(400 = 1600 - 2.5BC\)

\(2.5BC = 1200\)

\(BC = \frac{1200}{2.5} = 480 \text{ m}\)

(iii) The work done by the engine is:

\(30000 \times 14 = 420000 \text{ J}\)

The gain in kinetic energy is:

\(\frac{1}{2} \times 600 \times (30^2 - 20^2) = 150000 \text{ J}\)

The work done against resistance is:

\(750 \times CD = 420000 - 150000\)

\(750 \times CD = 270000\)

\(CD = \frac{270000}{750} = 360 \text{ m}\)

(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. Thus, \(F = \frac{P}{v} = \frac{720}{12} = 60 \, \text{N}\).

Using Newton's second law, \(F - R = ma\), where \(m = 75 \, \text{kg}\) and \(a = 0.16 \, \text{m/s}^2\). Therefore, \(60 - R = 75 \times 0.16\).

Solving for \(R\), we get \(R = 60 - 12 = 48\).

(ii) Given \(a > 0\), we have \(\frac{P}{v} - R = ma\). Since \(a > 0\), it implies \(\frac{P}{v} > R\).

Substituting \(R = 48\), we have \(\frac{720}{v} > 48\).

Solving for \(v\), \(v < \frac{720}{48} = 15\). Thus, the speed is less than \(15 \, \text{m/s}\).

Using the formula for power, \(P = Fv\), where \(F\) is the force exerted by the engine, we have:

\(\frac{P}{v} - R = ma\)

For \(v = 19\) m/s and \(a = 0.6\) m/s\(^{-2}\):

\(\frac{P}{19} - R = 1250 \times 0.6\)

For \(v = 30\) m/s and \(a = 0.16\) m/s\(^{-2}\):

\(\frac{P}{30} - R = 1250 \times 0.16\)

We have two equations:

1. \(\frac{P}{19} - R = 750\)
2. \(\frac{P}{30} - R = 200\)

To eliminate \(R\), subtract the second equation from the first:

\(\frac{P}{19} - \frac{P}{30} = 750 - 200\)

\(\frac{30P - 19P}{570} = 550\)

\(11P = 550 \times 570\)

\(P = 28500\)

Substitute \(P = 28500\) back into the first equation:

\(\frac{28500}{19} - R = 750\)

\(1500 - R = 750\)

\(R = 750\)

(i) The work done by the tension is given by the formula:

\(ext{Work Done} = T imes d imes ext{cos}( heta)\)

where \(T = 65 \text{ N}\), \(d = 76 \text{ m}\), and \(\theta = 5^{\circ}\).

\(ext{Work Done} = 65 imes 76 imes ext{cos}(5^{\circ})\)

\(ext{Work Done} = 4920 \text{ J}\)

(ii) The rate of working (power) is given by:

\(P = T imes v imes ext{cos}( heta)\)

where \(v = 1.5 \text{ m/s}\).

\(P = 65 imes 1.5 imes ext{cos}(5^{\circ})\)

\(P = 97.1 \text{ W}\)

(a) To find the power of the car's engine at point A, we first calculate the driving force using Newton's second law:

\(D - 500 = 1200 \times 0.8\)

\(D = 1460 \text{ N}\)

The power \(P\) is given by:

\(P = D \times v = 1460 \times 15 = 21900 \text{ W}\)

(b) To show that the distance AB is 1362.6 m, we first calculate the change in kinetic energy:

\(\text{Change in KE} = \frac{1}{2} \times 1200 \times 32^2 - \frac{1}{2} \times 1200 \times 15^2\)

\(= 614400 - 135000 = 479400 \text{ J}\)

The work done by the engine is:

\(\text{Work done} = 21900 \times 53 = 1160700 \text{ J}\)

Using the work-energy principle:

\(1160700 - 500 \times d = 479400\)

\(500d = 1160700 - 479400\)

\(500d = 681300\)

\(d = \frac{681300}{500} = 1362.6 \text{ m}\)

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

\(\text{Gain in PE} = mgh = 1250 \times 9.8 \times 1.54 = 19250 \text{ J}\)

The total work done by the crane is the sum of the gain in potential energy and the work done against resistance:

\(\text{Work done by crane} = 19250 + 5750 = 25000 \text{ J}\)

(ii) Using the formula for power:

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

Given \(P = 1250 \text{ W}\) and \(\Delta W = 25000 \text{ J}\), we solve for \(\Delta t\):

\(1250 = \frac{25000}{\Delta t}\)

\(\Delta t = \frac{25000}{1250} = 20 \text{ s}\)

(i) Using Newton's second law, the net force on the car is given by:

\(F = ma\)

where \(m = 700 \text{ kg}\) and \(a = 2 \text{ m s}^{-2}\).

The driving force \(DF\) minus the resistance force equals the net force:

\(DF - 600 = 700 \times 2\)

\(DF - 600 = 1400\)

\(DF = 1400 + 600 = 2000 \text{ N}\)

(ii) The power \(P\) is given by:

\(P = Fv\)

where \(F = 2000 \text{ N}\) and \(v = 15 \text{ m s}^{-1}\).

\(P = 2000 \times 15 = 30000 \text{ W}\)

Thus, the rate of working is 30000 W or 30 kW.

(i) Using the formula for power, \(P = Fv\), where \(F\) is the net force, we have:

\(24000 = (R - 1250 \times 0.32) \times 20\)

Solving for \(R\), we get \(R = 800\) N.

(ii) At point \(B\), using \(P = Fv\), we have:

\(24000/29.9 - 800 = 1250a\)

Solving for \(a\), we get \(a = 0.002\) m/s².

(iii) For \(v = 30\), \(24000/30 - 800 = 0\), so the car cannot accelerate further, hence cannot reach 30 m/s.

(iv) Since \(29.9 \leq v < 30\), the speed is approximately constant.

(v) The approximately constant speed is 30 m/s with a maximum error of 0.1 m/s, or 29.95 m/s with a maximum error of 0.05 m/s, or 29.9 m/s with a maximum error of 0.1 m/s.

(vi) (a) Using \(P = \frac{\Delta WD}{\Delta t}\), \(24 = \frac{1200}{T}\), solving gives \(T = 50\) s.

(vi) (b) Using \(s = vt\), \(s = 30 \times 50\) or \(29.95 \times 50\) or \(29.9 \times 50\), gives \(s = 1500\) m or 1495 m.

First, calculate the driving force (DF) using the power formula:

\(P = F \cdot v\), where \(P = 400 \text{ W}\) and \(v = 4 \text{ m s}^{-1}\).

\(F = \frac{P}{v} = \frac{400}{4} = 100 \text{ N}\).

For uphill motion, apply Newton's second law:

\(DF - 80g \sin 2^{\circ} = 80a\).

Substitute \(DF = 100 \text{ N}\) and \(g = 9.8 \text{ m s}^{-2}\):

\(100 - 80 \times 9.8 \times \sin 2^{\circ} = 80a\).

Calculate \(\sin 2^{\circ} \approx 0.0349\):

\(100 - 80 \times 9.8 \times 0.0349 = 80a\).

\(100 - 27.352 = 80a\).

\(72.648 = 80a\).

\(a = \frac{72.648}{80} \approx 0.9 \text{ m s}^{-2}\).

For downhill motion, apply Newton's second law:

\(DF + 80g \sin 2^{\circ} = 80a\).

\(100 + 80 \times 9.8 \times 0.0349 = 80a\).

\(100 + 27.352 = 80a\).

\(127.352 = 80a\).

\(a = \frac{127.352}{80} \approx 1.6 \text{ m s}^{-2}\).

Using Newton's second law, the net force on the car is given by:

\(F - R = ma\)

At point A, the force equation is:

\(F_A - 800 = 600a_A\)

The power of the engine is 40 kW, so:

\(F_A = \frac{40000}{25} = 1600 \text{ N}\)

Substituting into the force equation at A:

\(1600 - 800 = 600a_A\)

\(800 = 600a_A\)

\(a_A = \frac{800}{600} = \frac{4}{3} \text{ m s}^{-2}\)

At point B, the acceleration is half of that at A:

\(a_B = \frac{1}{2}a_A = \frac{2}{3} \text{ m s}^{-2}\)

Using the force equation at B:

\(\frac{40000}{v_B} - 800 = 600 \left(\frac{2}{3}\right)\)

\(\frac{40000}{v_B} - 800 = 400\)

\(\frac{40000}{v_B} = 1200\)

\(v_B = \frac{40000}{1200} = 33.3 \text{ m s}^{-1}\)

First, calculate the driving force (DF) using the power formula:

\(\text{Power} = \text{Force} \times \text{Velocity}\)

\(35000 = \text{DF} \times 16\)

\(\text{DF} = \frac{35000}{16} = 2187.5 \text{ N}\)

Apply Newton's second law to find the acceleration:

\(\text{Net force} = \text{mass} \times \text{acceleration}\)

The net force is given by:

\(\text{DF} - 1150g \sin(1.2^{\circ}) - 975 = 1150a\)

Substitute the known values:

\(2187.5 - 1150 \times 9.8 \times \sin(1.2^{\circ}) - 975 = 1150a\)

Calculate the gravitational component:

\(1150 \times 9.8 \times \sin(1.2^{\circ}) \approx 235.8 \text{ N}\)

Substitute back:

\(2187.5 - 235.8 - 975 = 1150a\)

\(976.7 = 1150a\)

Solve for \(a\):

\(a = \frac{976.7}{1150} \approx 0.849 \text{ m s}^{-2}\)

Rounding to three significant figures gives:

\(a = 0.845 \text{ m s}^{-2}\)

(i) At maximum speed, the driving force (DF) equals the resistance (R), so:

\(DF = R = 600 \text{ N}\)

The power (P) is given by:

\(P = DF \times v\)

Substituting the given power:

\(24000 = 600 \times v\)

Solving for \(v\):

\(v = \frac{24000}{600} = 40 \text{ m s}^{-1}\)

Thus, the speed cannot exceed 40 m s^-1.

(ii) Using Newton's second law, the net force is:

\(DF - R = ma\)

Substituting the given values:

\(\frac{24000}{15} - 600 = 1250a\)

Solving for \(a\):

\(1600 - 600 = 1250a\)

\(1000 = 1250a\)

\(a = \frac{1000}{1250} = 0.8 \text{ m s}^{-2}\)

Thus, the acceleration is 0.8 m s^-2.

(i) Using Newton's second law, the net force on the car is given by:

\(F - 900 = 1200a\)

where \(F\) is the force provided by the engine, and \(a\) is the acceleration (or deceleration in this case).

The power \(P\) of the engine is related to the force and speed by:

\(F = \frac{P}{v}\)

Substituting the given values:

\(F = \frac{18000}{25} = 720 \text{ N}\)

Substitute \(F\) back into the equation:

\(720 - 900 = 1200a\)

\(-180 = 1200a\)

\(a = -0.15 \text{ m s}^{-2}\)

Thus, the deceleration is 0.15 m s^-2.

(ii) To find the least speed, set the net force to zero:

\(\frac{18000}{v} - 900 = 0\)

Solve for \(v\):

\(\frac{18000}{v} = 900\)

\(v = \frac{18000}{900} = 20 \text{ m s}^{-1}\)

Therefore, the least speed is 20 m s^-1.

(i) The horizontal component of the tension is given by:

\(T \cos(20^{\circ}) = 851 \times \cos(20^{\circ})\)

The distance moved in 12 seconds is:

\(2.5 \times 12 = 30 \text{ m}\)

Therefore, the work done is:

\(\text{Work Done} = 851 \times 30 \times \cos(20^{\circ}) \approx 24000 \text{ J} = 24 \text{ kJ}\)

(ii) The power is given by:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{24000}{12} = 2000 \text{ W} = 2 \text{ kW}\)

First, apply Newton's second law to find the driving force \(DF\):

\(DF - 550 = 900 \times 0.2\)

\(DF = 180 + 550 = 730 \text{ N}\)

Next, use the formula for power:

\(P = \frac{(DF) \times v}{1000}\)

\(P = \frac{730 \times 30}{1000}\)

\(P = 21.9 \text{ kW}\)

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

\(\text{Power} = \text{Force} \times \text{Velocity} = 1400 \times 28\)

Calculating this gives:

\(\text{Power} = 39200 \text{ W} = 39.2 \text{ kW}\)

Rounding to 39.3 kW as per the mark scheme.

(b) The power equation is:

\(DF = \frac{43500}{v}\)

Resolving forces parallel to the hill:

\(DF = 1400 + 1250g \times 0.12 = 2900\)

Solving for speed:

\(v = \frac{43500}{2900} = 15 \text{ m s}^{-1}\)

(c) Using Newton's second law for the system:

For the car:

\(5000 - 1400 - 1250g \times 0.12 - T = 1250a\)

For the trailer:

\(T - 300 - 600g \times 0.12 = 600a\)

For the system:

\(5000 - 1400 - 300 - 1250g \times 0.12 - 600g \times 0.12 = (1250 + 600)a\)

Solving these equations gives:

\(a = 0.584 \text{ m s}^{-2}, \quad T = 1370 \text{ N}\)

(i) The power provided by the engine is 18 kW, which is 18000 W. At maximum speed, the power is used to overcome the resistance force R. The formula for power is given by:

\(P = F \cdot v\)

where \(P = 18000\) W and \(v = 30\) m/s. Solving for \(F\):

\(18000 = R \cdot 30\)

\(R = \frac{18000}{30} = 600\) N

(ii) Using Newton's second law, the net force \(F\) is given by:

\(F = ma\)

At 20 m/s, the power is still 18000 W, so the driving force \(DF\) is:

\(DF = \frac{18000}{20} = 900\) N

The net force is the driving force minus the resistance:

\(900 - 600 = 1200a\)

\(300 = 1200a\)

\(a = \frac{300}{1200} = 0.25\) m/s²

To find the acceleration, we use the formula for power:

\(P = Fv\)

where \(P\) is the power, \(F\) is the force, and \(v\) is the velocity.

Rearranging for force, we have:

\(F = \frac{P}{v} = \frac{420}{5} = 84 \text{ N}\)

(i) For the horizontal road:

Using Newton's second law, \(F = ma\), we have:

\(84 = 75a\)

\(a = \frac{84}{75} = 1.12 \text{ m/s}^2\)

(ii) For the inclined road:

The component of gravitational force along the slope is \(75g \sin(1.5^{\circ})\).

Using Newton's second law, \(F - 75g \sin(1.5^{\circ}) = 75a\), we have:

\(84 - 75 \times 9.8 \times \sin(1.5^{\circ}) = 75a\)

\(a = \frac{84 - 75 \times 9.8 \times \sin(1.5^{\circ})}{75}\)

\(a \approx 0.858 \text{ m/s}^2\)

First, apply Newton's second law along the direction of motion:

\(F - 1130 + 1250 \cdot g \cdot \sin(3^{\circ}) = 1250 \cdot 0.2\)

where \(F\) is the driving force, \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).

Rearranging gives:

\(F = 1130 - 1250 \cdot g \cdot \sin(3^{\circ}) + 250\)

Next, use the power equation:

\(P = Fv\)

Given \(P = 22000\) W, substitute for \(F\):

\(22000 = (1130 - 1250 \cdot g \cdot \sin(3^{\circ}) + 250) \cdot v\)

Solve for \(v\):

\(v = \frac{22000}{725.8} \approx 30.3 \text{ m/s}\)

(i) Using Newton's second law, the net force on the car is given by:

\(F = ma = 1200 \times 0.5 = 600 \text{ N}\)

The driving force (DF) minus the resistance is equal to the net force:

\(\text{DF} - 400 = 600\)

Thus, the driving force is:

\(\text{DF} = 1000 \text{ N}\)

The power is given by:

\(P = Fv\)

Substituting the values, we have:

\(20000 = 1000v\)

Solving for \(v\):

\(v = 20 \text{ m/s}\)

(ii) At maximum speed, the acceleration is zero, so the net force is zero:

\(\frac{20000}{v} - 400 = 0\)

Solving for \(v\):

\(\frac{20000}{v} = 400\)

\(v = 50 \text{ m/s}\)

(iii) The work done is given by:

\(\Delta W = 1500 \text{ kJ} = 1500000 \text{ J}\)

The distance \(d\) is:

\(d = \frac{1500000}{400} = 3750 \text{ m}\)

The time taken \(t\) is:

\(t = \frac{3750}{50} = 75 \text{ s}\)

To find the force produced by the engine, we use the formula for power:

\(P = F \times v\)

where \(P = 8000 \text{ W}\) (since 8 kW = 8000 W) and \(v = 25 \text{ m/s}\). Solving for \(F\):

\(F = \frac{P}{v} = \frac{8000}{25} = 320 \text{ N}\)

To find the resistance to motion, we apply Newton's second law:

\(F - R = ma\)

where \(F = 320 \text{ N}\), \(m = 100 \text{ kg}\), and \(a = 0.5 \text{ m/s}^2\). Solving for \(R\):

\(320 - R = 100 \times 0.5\)

\(320 - R = 50\)

\(R = 320 - 50 = 270 \text{ N}\)

(i) Since the crate is lifted at constant speed, the tension in the cable equals the weight of the crate. The weight is given by:

\(W = mg\)

where \(m = 800 \text{ kg}\) and \(g = 10 \text{ m/s}^2\) (standard gravity approximation). Thus,

\(W = 800 \times 10 = 8000 \text{ N}\)

Alternatively, using \(g = 9.8 \text{ m/s}^2\) or \(g = 9.81 \text{ m/s}^2\), the tension can be \(7840 \text{ N}\) or \(7850 \text{ N}\) respectively.

(ii) The power applied is calculated using the formula:

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

where \(\Delta W = 800 \times 20\) (work done to lift the crate by 20 m) and \(\Delta t = 50 \text{ s}\).

\(\Delta W = 16000 \text{ J}\)

\(P = \frac{16000}{50} = 320 \text{ W}\)

Alternatively, using \(g = 9.8 \text{ m/s}^2\) or \(g = 9.81 \text{ m/s}^2\), the power can be \(3140 \text{ W}\).

First, calculate the driving force using the power formula:

\(Power = Force × Velocity\)

\(Given power = 20,000 W and velocity = 25 m/s,\)

Driving force = \(\frac{20,000}{25} = 800 \text{ N}\).

Using Newton's second law, the net force is given by:

\(Net force = Driving force - Resistance = 800 - 600 = 200 \text{ N}.\)

According to Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.

Substitute the known values:

\(200 = 1000a\)

\(a = \frac{200}{1000} = 0.2 \text{ m s}^{-2}\).

(a) The forward force exerted by the cyclist's driving force is given by \(\frac{180}{6} = 30 \text{ N}\).

Using Newton's second law, the equation of motion is:

\(30 - F - 70g \sin \alpha = 70 \times (-0.2)\)

Substitute \(g = 9.8 \text{ m s}^{-2}\) and \(\sin \alpha = 0.05\):

\(30 - F - 70 \times 9.8 \times 0.05 = 70 \times (-0.2)\)

\(30 - F - 34.3 = -14\)

\(F = 9 \text{ N}\)

(b) For steady speed, acceleration \(a = 0\). Using the power equation:

\(\frac{180}{v} - F - 70g \sin \alpha = 0\)

Substitute \(F = 9 \text{ N}\) and \(\sin \alpha = 0.05\):

\(\frac{180}{v} - 9 - 34.3 = 0\)

\(\frac{180}{v} = 43.3\)

\(v = \frac{180}{43.3} \approx 4.09 \text{ m s}^{-1}\)