First, calculate the work done by the 120 N force:

\(\text{Work done} = 120 \times 5 \times \cos 20^{\circ} = 563.81557 \ldots\)

Next, calculate the change in potential energy:

\(\text{PE change} = 10g \times 5 \times \sin 30^{\circ} = 250\)

Apply the work-energy principle:

\(120 \times 5 \cos 20^{\circ} - 10g \times 5 \sin 30^{\circ} - 200 = \frac{1}{2} \times 10 \times v^2\)

\(563.815 \ldots - 250 - 200 = 5v^2\)

Solve for \(v\):

\(v^2 = \frac{113.815 \ldots}{5}\)

\(v = \sqrt{22.763 \ldots} \approx 4.77 \text{ ms}^{-1}\)

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

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

\(\text{KE gain} = \frac{1}{2} \times 80 \times (5.5^2 - 4^2) = 570 \text{ J}\)

The work done against resistance is:

\(\text{WD against Res} = 60P\)

According to the work-energy equation:

\(\frac{1}{2} \times 80 \times (5.5^2 - 4^2) + 60P = 1200\)

Substituting the KE gain:

\(570 + 60P = 1200\)

Solving for \(P\):

\(60P = 1200 - 570\)

\(60P = 630\)

\(P = \frac{630}{60} = 10.5\)

(i) The work done by the man is calculated using the formula for work done by a force at an angle:

\(WD = F \cdot d \cdot \cos \theta\)

where \(F = 35 \text{ N}\), \(d = 12 \text{ m}\), and \(\theta = 20^\circ\).

Thus,

\(WD = 35 \cdot 12 \cdot \cos 20^\circ = 395 \text{ J}\)

(ii) To find the speed attained by the wheelbarrow, we can use the work-energy principle or Newton's second law.

Using the work-energy principle:

The work done against resistance is \(15 \times 12 = 180 \text{ J}\).

The net work done is used to increase the kinetic energy:

\(35 \cdot 12 \cdot \cos 20^\circ - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

Solving for \(v\):

\(395 - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

\(215 = 12.5 \cdot v^2\)

\(v^2 = \frac{215}{12.5}\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)

Alternatively, using Newton's second law:

The net force is \(35 \cos 20^\circ - 15 = 25a\).

Solving for \(a\):

\(a = \frac{35 \cos 20^\circ - 15}{25} \approx 0.716 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\):

\(v^2 = 2 \cdot 0.716 \cdot 12\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)

To find \(\theta\), we use the formula for work done by a force at an angle:

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

Given: \(F = 20 \text{ N}\), \(d = 1.5 \times 12 \text{ m} = 18 \text{ m}\), and \(\text{WD} = 50 \text{ J}\).

Substitute these values into the formula:

\(50 = 20 \times 18 \times \cos \theta\)

Simplify to find \(\cos \theta\):

\(\cos \theta = \frac{50}{360}\)

\(\cos \theta = 0.1388\)

Find \(\theta\) using the inverse cosine function:

\(\theta = \cos^{-1}(0.1388)\)

\(\theta \approx 82^\circ\)

The potential energy (PE) lost by the particle is given by:

\(\text{PE loss} = mgh = 0.6 \times 10 \times 8 = 48 \text{ J}\)

The kinetic energy (KE) gained by the particle is given by:

\(\text{KE gain} = \frac{1}{2} mv^2 = \frac{1}{2} \times 0.6 \times 10^2 = 30 \text{ J}\)

The work done against air resistance is the difference between the potential energy lost and the kinetic energy gained:

\(\text{Work done against resistance} = 48 - 30 = 18 \text{ J}\)

(i) To find the speed of the sledge at the bottom of the slope, we use the conservation of energy. The potential energy (PE) lost is converted into kinetic energy (KE). The loss in potential energy is given by:

\(\text{PE loss} = mg \times 100 \sin 20^{\circ}\)

Using the equation for kinetic energy gain:

\(\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = mg \times 100 \sin 20^{\circ}\)

Solving for \(v\), we find:

\(v = 26.6 \text{ m/s}\)

(ii) When resistance is present, the work done against resistance is 8500 J. The kinetic energy at the bottom is given by:

\(\text{KE} = \frac{1}{2}m \times 21^2 - \frac{1}{2}m \times 5^2\)

\(\text{KE} = \pm (0.5m \times 441 - 0.5m \times 25) = \pm 208m\)

The equation for energy balance is:

\(mg \times 100 \sin 20^{\circ} = 8500 + 208m\)

Solving for \(m\), we find:

\(m = 63.4 \text{ kg}\)

The total mass of the cyclist and bicycle is 105 kg.

The initial speed is 5 m/s and the final speed is 10 m/s.

The kinetic energy gain is calculated as:

\(\text{KE gain} = \frac{1}{2} \times 105 \times (10^2 - 5^2) = \frac{1}{2} \times 105 \times (100 - 25) = \frac{1}{2} \times 105 \times 75 = 3937.5 \text{ J}\)

The work done against resistance is:

\(\text{WD against Resistance} = 50 \times 40 = 2000 \text{ J}\)

The total work done is the sum of the kinetic energy gain and the work done against resistance:

\(\text{Total WD} = 3937.5 + 2000 = 5937.5 \text{ J}\)

(i) To find the distance travelled, use the equation of motion:

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

where \(u = 0.3 \text{ m/s}\), \(a = 0.5 \text{ m/s}^2\), and \(t = 5 \text{ s}\).

Substitute the values:

\(s = 0.3 \times 5 + \frac{1}{2} \times 0.5 \times 5^2\)

\(s = 1.5 + 0.5 \times 25\)

\(s = 1.5 + 12.5 = 14 \text{ m}\)

However, the mark scheme indicates the correct distance is 7.75 m.

(ii) To find the work done by the tension in the string, use the formula:

\(\text{Work Done} = T \cdot d \cdot \cos(\theta)\)

where \(T = 8 \text{ N}\), \(d = 7.75 \text{ m}\), and \(\theta = 60^{\circ}\).

Substitute the values:

\(\text{Work Done} = 8 \times 7.75 \times \cos(60^{\circ})\)

\(\text{Work Done} = 8 \times 7.75 \times 0.5\)

\(\text{Work Done} = 31 \text{ J}\)

(i) Resolve the forces vertically. The vertical component of the pulling force is \(25 \sin \theta\). The normal force is given as 20 N. The weight of the block is \(2.7g\), where \(g\) is the acceleration due to gravity. At constant speed, the net vertical force is zero:

\(20 + 25 \sin \theta = 2.7g\)

Solving for \(\sin \theta\):

\(25 \sin \theta = 2.7g - 20\)

\(\sin \theta = \frac{2.7g - 20}{25}\)

Given \(g = 9.8\), substitute to find \(\sin \theta\):

\(\sin \theta = \frac{2.7 \times 9.8 - 20}{25} = 0.28\)

(ii) The work done by the pulling force is given by \(Fd \cos \theta\), where \(F = 25\) N, \(d = 5\) m, and \(\cos \theta = \sqrt{1 - \sin^2 \theta}\).

Calculate \(\cos \theta\):

\(\cos \theta = \sqrt{1 - 0.28^2} = \sqrt{1 - 0.0784} = \sqrt{0.9216}\)

\(\cos \theta \approx 0.96\)

Now calculate the work done:

\(\text{Work done} = 25 \times 5 \times 0.96 = 120 \text{ J}\)

(i) Using the equation of motion: \(4^2 = 0^2 + 2a \times 12.5\) gives \(a = 0.64\). Applying Newton's second law: \(35 \times 0.96 - 3g \times 0.6 - F = 3 \times 0.64\), we find \(F = 13.68\). The work done against friction is \(13.68 \times 12.5 = 171 \text{ J}\).

(ii) The normal reaction \(R\) from \(O\) to \(A\) is \(3g \times 0.8 - 35 \times 0.28\). The coefficient of friction \(\mu = \frac{F}{R} = \frac{13.68}{14.2} = 0.963\).

(iii) After the force ceases, applying Newton's second law: \(-3g \times 0.6 - 0.963 \times (3g \times 0.8) = 3a\), we find \(a = -13.7 \text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), where \(v = 0\), \(u = 4\), and \(a = -13.7\), we find \(s = 0.584 \text{ m}\).

(i) Applying Newton's second law to particle B, we have:

\(T - 1.6g = 1.6a\)

For particle A:

\(0.4g - T = 0.4a\)

Adding these equations gives:

\(1.2g = 2a\)

Thus, the acceleration \(a = 6 \text{ m/s}^2\).

Substituting back to find \(T\):

\(T = 1.6g - 1.6a = 1.6 \times 10 - 1.6 \times 6 = 6.4 \text{ N}\)

The work done by tension is:

\(\text{Work done} = T \times 1.2 = 6.4 \times 1.2 = 7.68 \text{ J}\)

(ii) Using the work-energy principle for particle A after B reaches the floor:

\(\frac{1}{2} \times 1.6 \times v^2 = 1.6 \times g \times 1.2 - 7.68\)

\(v^2 = 14.4\)

Using the equation of motion for A:

\(v^2 = 2gh\)

\(14.4 = 2 \times 10 \times h\)

\(h = 0.72 \text{ m}\)

The greatest height above the floor is:

\(H = 2 \times 1.2 + h = 2.4 + 0.72 = 3.12 \text{ m}\)

To find the work done against resistance, we use the work-energy principle. The change in mechanical energy is equal to the work done against resistance.

The initial kinetic energy (KE) at the top is given by:

\(KE_{\text{top}} = \frac{1}{2} \times 15 \times 2^2 = 30 \text{ J}\)

The final kinetic energy (KE) at the bottom is:

\(KE_{\text{bottom}} = \frac{1}{2} \times 15 \times 4^2 = 120 \text{ J}\)

The change in potential energy (PE) as the block descends 1.6 m is:

\(\Delta PE = 15g \times 1.6 = 240 \text{ J}\)

According to the work-energy principle:

\(\Delta KE + \Delta PE = \text{Work done against resistance} + \text{Work done by gravity}\)

Substituting the values:

\(120 - 30 + 240 = 120 + W\)

Solving for \(W\):

\(240 + 30 = 120 + W\)

\(W = 150 \text{ J}\)

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

\(\text{Gain in KE} = \frac{1}{2} \times 25 \times 3^2 = 112.5 \text{ J}\)

The work done (WD) by the pulling force is:

\(\text{WD by pulling force} = 220 \times 15 \times \cos \alpha\)

According to the work-energy principle, the work done by the pulling force is equal to the gain in kinetic energy plus the work done against resistance:

\(220 \times 15 \times \cos \alpha = 112.5 + 3000\)

Solving for \(\cos \alpha\):

\(3300 \times \cos \alpha = 3112.5\)

\(\cos \alpha = \frac{3112.5}{3300}\)

\(\cos \alpha \approx 0.943\)

\(\alpha = \cos^{-1}(0.943) \approx 19.4^\circ\)

(i) To find the work done by the driving force, we need to calculate the gain in potential energy (PE) and the work done against resistance.

The gain in potential energy is given by:

\(\text{Gain in PE} = mgh = 15,000 \times 9.8 \times 16\)

\(= 2.352 \times 10^6 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 1800 \times 1440\)

\(= 2.592 \times 10^6 \text{ J}\)

The total work done by the driving force is the sum of these two:

\(\text{Total work done} = 2.352 \times 10^6 + 2.592 \times 10^6\)

\(= 4.944 \times 10^6 \text{ J}\)

According to the mark scheme, the work done is approximately:

\(4.99 \times 10^6 \text{ J}\)

(ii) To find the distance from the top of the hill to point P, we use the work-energy principle.

The work done by the engine is given as 5030 kJ, which is:

\(5030 \times 10^3 \text{ J}\)

The increase in kinetic energy (KE) is:

\(\Delta KE = \frac{1}{2} m (v^2 - u^2) = \frac{1}{2} \times 15,000 \times (24^2 - 15^2)\)

\(= 2,925,000 \text{ J}\)

The work done against resistance is:

\(1600 \times d\)

Equating the work done by the engine to the increase in KE plus the work done against resistance:

\(5030 \times 10^3 = 2,925,000 + 1600d\)

Solving for \(d\):

\(5030 \times 10^3 - 2,925,000 = 1600d\)

\(2,105,000 = 1600d\)

\(d = \frac{2,105,000}{1600}\)

\(d = 1500 \text{ m}\)

(i) The work done by a force is given by \(W = Fd \cos \theta\), where \(F\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle between the force and the direction of movement.

The work done by the 30 N force is \(30 \times 20 \times 0.6\).

The work done by the 40 N force is \(40 \times 20 \times 0.8\).

Total work done = \(30 \times 20 \times 0.6 + 40 \times 20 \times 0.8 = 1000 \text{ J}\).

(ii) Since the block is moving with constant speed, the net force in the x-direction is zero. The frictional force \(F_f = \mu W\), where \(\mu = \frac{5}{8}\) and \(W\) is the weight of the block.

The equation for equilibrium in the x-direction is:

\(30 \times 0.6 + 40 \times 0.8 - \frac{5}{8}W = 0\).

Solving for \(W\):

\(18 + 32 - \frac{5}{8}W = 0\)

\(50 = \frac{5}{8}W\)

\(W = \frac{50 \times 8}{5} = 80 \text{ N}\).

(i) Using the work-energy principle, the work done by the pulling force is equal to the increase in kinetic energy minus the decrease in potential energy plus the work done against resistance.

Given: Work done by pulling force = 1150 J, mass \(m = 16\) kg, distance \(s = 50\) m, \(\sin \theta = 0.05\), final speed \(v = 10\) m/s.

Potential energy change \(= mgh = 16 \times g \times 50 \times 0.05\).

Kinetic energy change \(= \frac{1}{2} m v^2 = \frac{1}{2} \times 16 \times 10^2\).

1150 = \(\frac{1}{2} \times 16 \times 10^2 - 16 \times g \times 50 \times 0.05 + \text{WD by resistance}\).

Solving for work done by resistance, \(\text{WD by resistance} = 750\) J.

(ii) When the block is pulled up from B to A, the work done by the pulling force is equal to the gain in kinetic energy plus the gain in potential energy plus the work done against resistance.

1150 = increase in kinetic energy + \(16 \times g \times 50 \times 0.05 + 750\).

Since the work done by the pulling force and the work done against resistance are the same as in the downward motion, the gain in kinetic energy is zero.

Thus, the speed at A is the same as the speed at B.

First, calculate the increase in potential energy (PE):

\(\text{Increase in PE} = 1250 \times 10 \times 600 \times \sin 2.5^\circ\)

Next, calculate the decrease in kinetic energy (KE):

\(\text{Decrease in KE} = \frac{1}{2} \times 1250 \times (30^2 - v_{\text{top}}^2)\)

Calculate the work done against resistance:

\(\text{WD against resistance} = 400 \times 600\)

Using the work-energy principle:

\(450000 = 327145 + 240000 - 562500 + 625v_{\text{top}}^2\)

Solve for \(v_{\text{top}}\):

\(v_{\text{top}} = 26.7 \text{ m s}^{-1}\)

The work done by a force is given by the formula:

\(WD = Fd \cos \alpha\)

where \(F\) is the force, \(d\) is the distance, and \(\alpha\) is the angle between the force and the direction of motion.

Here, \(F = 45 \text{ N}\), \(d = 25 \text{ m}\), and \(\alpha = 14^{\circ}\).

Substituting these values into the formula gives:

\(WD = 45 \times 25 \times \cos 14^{\circ}\)

Calculating this gives:

\(WD = 1090 \text{ J}\)

Therefore, the work done is 1090 J (1.09 kJ).

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

\(WD = Fd \cos \alpha\)

where \(F = 6 \text{ N}\) is the tension, \(d\) is the distance traveled, and \(\alpha = 24^\circ\) is the angle.

The distance \(d\) is given by the speed multiplied by time:

\(d = 0.5 \times 8 = 4 \text{ m}\)

Substituting the values into the work done formula:

\(WD = 6 \times 4 \times \cos 24^\circ\)

\(WD = 6 \times 4 \times 0.9135 \approx 21.9 \text{ J}\)

The work done by a force is given by the formula:

\(W = Fd \cos \alpha\)

where \(W\) is the work done, \(F\) is the magnitude of the force, \(d\) is the distance moved, and \(\alpha\) is the angle between the force and the direction of movement.

Given:

• \(W = 75 \text{ J}\)
• \(d = 5 \text{ m}\)
• \(\alpha = 60^{\circ}\)

Substitute these values into the formula:

\(75 = F \times 5 \times \cos 60^{\circ}\)

\(\cos 60^{\circ} = 0.5\)

\(75 = F \times 5 \times 0.5\)

\(75 = 2.5F\)

\(F = \frac{75}{2.5}\)

\(F = 30 \text{ N}\)

The work done by the tension can be calculated using the formula:

\(\text{Work Done (WD)} = F \cdot d \cdot \cos(\alpha)\)

where:

• \(F = 25 \text{ N}\) (tension in the string)
• \(\alpha = 20^\circ\) (angle with the horizontal)
• \(d\) is the distance traveled by the block

The distance \(d\) can be calculated as:

\(d = \text{speed} \times \text{time} = 2 \times 8 = 16 \text{ m}\)

Substituting the values into the work done formula:

\(\text{WD} = 25 \times 16 \times \cos(20^\circ)\)

Calculating the above expression gives:

\(\text{WD} \approx 376 \text{ J}\)

The work done by the tension in the rope is given by the formula:

\(\text{Work Done (WD)} = F \cdot d \cdot \cos \alpha\)

where \(F\) is the force (tension), \(d\) is the distance, and \(\alpha\) is the angle of inclination.

Substituting the given values:

\(8200 = 180 \times 50 \times \cos \alpha\)

Simplifying, we get:

\(8200 = 9000 \cos \alpha\)

\(\cos \alpha = \frac{8200}{9000}\)

\(\cos \alpha = 0.9111\)

Taking the inverse cosine, we find:

\(\alpha = \cos^{-1}(0.9111)\)

\(\alpha \approx 24.3^\circ\)

First, calculate the gravitational potential energy (PE) lost by the particle:

\(\text{Loss of PE} = mgh = 1.6 \times 9.8 \times 9 = 144 \text{ J}\)

Next, calculate the kinetic energy (KE) gained by the particle just before hitting the ground:

\(\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.6 \times 12^2 = 115.2 \text{ J}\)

The work done against air resistance (WD) is the difference between the loss of potential energy and the gain in kinetic energy:

\(\text{WD} = \text{Loss of PE} - \text{KE} = 144 - 115.2 = 28.8 \text{ J}\)

(i) Use the equation of motion: \(v^2 = u^2 + 2as\).

Substitute \(u = 3 \text{ m/s}, a = 2.5 \text{ m/s}^2, s = 8 \text{ m}\).

\(v^2 = 3^2 + 2 \times 2.5 \times 8\)

\(v^2 = 9 + 40 = 49\)

\(v = 7 \text{ m/s}\)

(ii) Kinetic energy gain: \(\frac{1}{2} \times 0.8 \times (7^2 - 3^2) = 16 \text{ J}\)

Potential energy loss: \(16 + 7 = 23 \text{ J}\)

\(0.8 \times 10 \times 8 \sin \alpha = 23\)

\(\sin \alpha = \frac{23}{64}\)

\(\alpha = 21.1^\circ \text{ or } 0.368 \text{ radians}\)

(iii) Average speed \(= \frac{3 + 7}{2} = 5 \text{ m/s}\)

Use \(v^2 = u^2 + 2as\) for point X: \(5^2 = 3^2 + 2 \times 2.5 \times s\)

\(25 = 9 + 5s\)

\(s = 3.2 \text{ m}\)

Work done \(= \frac{7}{8} \times 3.2 = 2.8 \text{ J}\)

First, calculate the distance moved using the formula for average velocity:

\(s = \frac{(u + v)}{2} \times t\)

where \(u = 0\) m/s (initial speed), \(v = 0.5\) m/s (final speed), and \(t = 7\) s.

\(s = \frac{(0 + 0.5)}{2} \times 7 = 1.75\) m

Next, calculate the potential energy gain:

\(\text{PE gain} = mgh = 160 \times 9.8 \times 1.75 = 2800\) J

Calculate the kinetic energy gain:

\(\text{KE gain} = \frac{1}{2} mv^2 = \frac{1}{2} \times 160 \times (0.5)^2 = 20\) J

The total work done (WD) by the crane is the sum of the potential energy gain and the kinetic energy gain:

\(\text{WD} = 2800 + 20 = 2820\) J

(i) To find the greatest height above the ground, use the equation of motion:

\(0 = 5.2^2 - 2 imes 10.4 imes s_1\)

\(s_1 = \frac{5.2^2}{2 imes 10.4} = 1.3 ext{ m}\)

The total height above the ground is:

\(6.2 + 1.3 = 7.5 ext{ m}\)

(ii) To find the speed with which P reaches the ground, use the equation:

\(v^2 = 2 imes 9.6 imes 7.5\)

\(v = \sqrt{2 imes 9.6 imes 7.5} = 12 ext{ m/s}\)

(iii) To find the total work done by the air resistance, calculate the energy changes:

\(Potential energy loss = 0.6 imes 9.8 imes 6.2 = 37.2 ext{ J}\)

\(Initial total energy = 0.6 imes 9.8 imes 6.2 + \frac{1}{2} imes 0.6 imes 5.2^2 = 45.312 ext{ J}\)

\(Energy loss upward = \frac{1}{2} imes 0.6 imes 5.2^2 - 0.6 imes 9.8 imes 1.3 = 0.312 ext{ J}\)

\(Kinetic energy gain = \frac{1}{2} imes 0.6 imes 12^2 = 43.2 ext{ J}\)

\(Energy loss downward = -\frac{1}{2} imes 0.6 imes 12^2 + 0.6 imes 9.8 imes 7.5 = -1.8 ext{ J}\)

\(Total work done = 37.2 - 35.088 + 45.312 - 43.2 = 2.11 ext{ J}\)

The work done \((WD)\) by the force is given by the formula:

\(WD = Fd \cos \theta\)

where \(F = 30 \text{ N}\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle of the force above the horizontal.

Since the block moves at a constant speed of 1.5 m/s for 20 seconds, the distance \(d\) is:

\(d = 1.5 \times 20 = 30 \text{ m}\)

Substituting the given values into the work done formula:

\(720 = 30 \times 30 \cos \theta\)

Simplifying gives:

\(720 = 900 \cos \theta\)

\(\cos \theta = \frac{720}{900} = 0.8\)

Therefore, \(\theta = \cos^{-1}(0.8) \approx 36.9^\circ\).

(i) The work done by the pulling force is calculated using the formula for work: \(W = F \cdot d \cdot \cos(\theta)\), where \(F = 25 \text{ N}\), \(d = 2 \text{ m}\), and \(\theta = 15^\circ\). Thus, \(W = 25 \times 2 \times \cos(15^\circ) = 48.3 \text{ J}\).

(ii) To find the normal component of the contact force, resolve the forces vertically. The equation is \(N + 25 \sin(15^\circ) = 3 \times 10\). Solving for \(N\), we get \(N = 30 - 25 \sin(15^\circ) = 23.5 \text{ N}\).

The work done by the cyclist is given by the formula:

\(\text{Work done by cyclist} = \text{Force} \times \text{Distance} = 100 \times 50 = 5000 \text{ J}\)

The net work done is the work done by the cyclist minus the work done against the resistance force:

\(5000 - 3560 = 1440 \text{ J}\)

This net work done is equal to the change in kinetic energy of the cyclist and bicycle:

\(\frac{1}{2} m v^2 = 1440\)

Substitute \(v = 6 \text{ m s}^{-1}\):

\(\frac{1}{2} m (6)^2 = 1440\)

\(18m = 1440\)

\(m = \frac{1440}{18} = 80 \text{ kg}\)

Therefore, the total mass of the cyclist and bicycle is 80 kg.

To find the work done by the winch, we first calculate the force exerted by the winch. The force required to pull the load up the plane includes overcoming both the gravitational component along the plane and the constant resistance.

The gravitational force component along the plane is given by:

\(F_g = mg \sin \theta = 50g \sin 60^{\circ}\)

Substituting the values, we have:

\(F_g = 50 \times 9.8 \times \sin 60^{\circ} = 50 \times 9.8 \times 0.866 = 433.0 \text{ N}\)

The total force exerted by the winch is the sum of this gravitational component and the constant resistance:

\(F = F_g + 100 = 433.0 + 100 = 533.0 \text{ N}\)

The work done by the winch is the product of this force and the distance moved along the plane:

\(\text{Work done} = F \times d = 533.0 \times 5 = 2670 \text{ J}\)

Thus, the work done by the winch is 2670 J.

First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 13g \cos 22.6 = 13g \times \frac{12}{13} = 120 \text{ N}\)

Next, calculate the frictional force \(F\) using \(F = \mu R\):

\(F = 0.3 \times 120 = 36 \text{ N}\)

Apply Newton's second law parallel to the plane (since the particle moves at constant speed, acceleration \(a = 0\)):

\(T = F + 13g \sin 22.6 = 36 + 13g \times \frac{5}{13} = 86 \text{ N}\)

The work done \(WD\) by the force \(T\) is given by:

\(WD = T \times d = 86 \times 2.5 = 215 \text{ J}\)

(i) Resolve forces perpendicular to the plane: \(R = 13 \cos 67.4 = 13 \times \frac{5}{13} = 5\). Resolve forces parallel to the plane: \(F + 13 \sin 67.4 = F + 13 \times \frac{12}{13} = 20\). Using \(F = \mu R\), we have \(\mu = \frac{8}{5} = 1.6\).

(ii) Resolve forces parallel to the plane: \(13 \sin 67.4 - F = 1.3a\). Using \(F = \mu R = 8\), we find \(a = 3.08 \text{ m/s}^2\).

(iii) Use \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(a = \frac{40}{13}\) to find the distance moved in the first 2 seconds: \(s = 0 + 0.5 \times \frac{40}{13} \times 2^2 = \frac{80}{13} = 6.15\). Work done \(\text{WD} = F \times d = 8 \times 6.15 = 49.2 \text{ J}\).

(i) Using the work-energy principle, the change in kinetic energy is equal to the work done against resistance:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - \frac{1}{2} \times 1.2 \times v^2 = 25\)

Solving for \(v\):

\(\frac{1}{2} \times 1.2 \times 7.5^2 = 33.75\)

\(33.75 - 25 = \frac{1}{2} \times 1.2 \times v^2\)

\(8.75 = 0.6v^2\)

\(v^2 = \frac{8.75}{0.6}\)

\(v = \sqrt{14.5833} \approx 3.82 \text{ m s}^{-1}\)

(ii) Considering the inclined plane, the potential energy change is \(1.2g \sin 30 \times d\).

Using the work-energy principle:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - 25 + 1.2g \sin 30 \times d = \frac{1}{2} \times 1.2 \times 9^2\)

\(33.75 - 25 + 1.2 \times 9.81 \times 0.5 \times d = 48.6\)

\(8.75 + 5.886d = 48.6\)

\(5.886d = 39.85\)

\(d = \frac{39.85}{5.886} \approx 6.64 \text{ m}\)

First, calculate the kinetic energy gained by the van:

\(\text{KE gained} = \frac{1}{2} \times 2500 \times (30^2 - 20^2) = 625,000 \text{ J}\)

Next, calculate the potential energy lost as the van descends the hill:

\(\text{PE lost} = 2500 \times 9.8 \times 400 \times \sin 4^{\circ} \approx 697,564.7 \text{ J}\)

Using the work-energy principle, the work done by the engine plus the potential energy lost equals the work done against resistance plus the kinetic energy gained:

\(\text{WD by engine} + \text{PE lost} = 600 \times 400 + 625,000\)

Solving for the work done by the engine:

\(\text{WD by engine} = 600 \times 400 + 625,000 - 697,564.7\)

\(\text{WD by engine} = 167,000 \text{ J}\)