1. Kinematic Relationship

Let $x$ be the horizontal position of the car from point A, and $h$ be the constant height AB. The length of the rope segment from the car to pulley B is $s = \sqrt{x^2+h^2}$. Let $y_D$ be the vertical position of object D, measured upwards from a fixed datum. The total length of the active rope is constant: $L = s + y_D = \text{constant}$.

We relate the velocities by differentiating with respect to time:

$$\dot{L} = \dot{s} + \dot{y}_D = 0$$

The upward velocity of object D is $v_{D,up} = \dot{y}_D$. Thus, $v_{D,up} = -\dot{s}$. The rate of change of the rope length $s$ is:

$$\dot{s} = \frac{d}{dt}\sqrt{x^2+h^2} = \frac{x\dot{x}}{\sqrt{x^2+h^2}}$$

Recognizing that $v_{car} = \dot{x}$ and from the geometry $\sin\phi = \frac{x}{\sqrt{x^2+h^2}}$, we get:

$$v_{D,up} = \dot{s} = v_{car}\sin\phi$$

Note: The original relation was $\dot{s} + \dot{y}_D = 0$. So $\dot{y}_D = -\dot{s} = -v_{car}\sin\phi$. This negative sign indicates that as the car moves right ($v_{car}>0$), object D moves down ($\dot{y}_D<0$). This seems incorrect, as pulling the rope should lift the object. Let's redefine $y_D$ as the length of the rope from pulley C to object D. Then $L = \sqrt{x^2+h^2} + y_D = \text{constant}$. Now $\dot{s} + \dot{y}_D = 0$, where $\dot{y}_D$ is the velocity of D downwards. So $v_{D,down} = \dot{y}_D = -\dot{s} = -v_{car}\sin\phi$. The upward velocity is $v_{D,up} = -v_{D,down} = v_{car}\sin\phi$. This is physically correct.

To find the acceleration, we differentiate the upward velocity of D:

$$a_{D,up} = \frac{d}{dt}(v_{D,up}) = \frac{d}{dt}(v_{car}\sin\phi)$$

Using the product rule for differentiation:

$$a_{D,up} = \dot{v}_{car}\sin\phi + v_{car}\cos\phi\dot{\phi}$$

The car has constant acceleration $a = \dot{v}_{car}$. To find $\dot{\phi}$, we use the geometric relation $x = h\tan\phi$. Differentiating with respect to time:

$\dot{x} = v_{car} = h\sec^2\phi\dot{\phi} \implies \dot{\phi} = \frac{v_{car}}{h}\cos^2\phi$.

Substituting $\dot{\phi}$ into the acceleration expression:

$$a_{D,up} = a\sin\phi + v_{car}\cos\phi\left(\frac{v_{car}}{h}\cos^2\phi\right)$$ $$a_{D,up} = a\sin\phi + \frac{v_{car}^2}{h}\cos^3\phi$$

2. State of the Car at $\phi=60^\circ$

Given the height $h = AB = 3\sqrt{3}$ m, we find the car's position $x$ at $\phi=60^\circ$:

$$x = h\tan\phi = (3\sqrt{3})\tan(60^\circ) = (3\sqrt{3})(\sqrt{3}) = 9 \text{ m}$$

The car starts from rest and moves with constant acceleration $a=2 \text{ m/s}^2$. We find its velocity squared using the kinematic equation $v^2 = u^2 + 2as$:

$$v_{car}^2 = 0 + 2(2)(9) = 36 \text{ (m/s)}^2$$

3. Acceleration of Object D

Now we substitute the known values for the car's state and the angle into the derived expression for the upward acceleration of object D, $a_{D,up}$:

$$a_{D,up} = (2)\sin(60^\circ) + \frac{36}{3\sqrt{3}}\cos^3(60^\circ)$$ $$a_{D,up} = 2\left(\frac{\sqrt{3}}{2}\right) + \frac{12}{\sqrt{3}}\left(\frac{1}{2}\right)^3$$ $$a_{D,up} = \sqrt{3} + \frac{12}{8\sqrt{3}} = \sqrt{3} + \frac{3}{2\sqrt{3}} = \sqrt{3} + \frac{\sqrt{3}}{2}$$ $$a_{D,up} = \frac{3\sqrt{3}}{2} \text{ m/s}^2$$

4. Tension in the Rope

Applying Newton's second law to object D, with the upward direction as positive:

$$\sum F_y = T - P = m_D a_{D,up}$$

where $T$ is the tension and $P$ is the weight of D. The mass of D is $m_D = P/g$.

$$T = P + m_D a_{D,up} = P + \frac{P}{g}a_{D,up}$$ $$T = P\left(1 + \frac{a_{D,up}}{g}\right)$$

Substituting the values $P=5000 \text{ N}$ and $a_{D,up} = \frac{3\sqrt{3}}{2} \text{ m/s}^2$, and using the standard gravitational acceleration $g=9.8 \text{ m/s}^2$:

$$T = 5000\left(1 + \frac{3\sqrt{3}/2}{9.8}\right) = 5000\left(1 + \frac{2.598}{9.8}\right)$$ $$T = 5000(1 + 0.2651) = 5000(1.2651) \approx 6325.5 \text{ N}$$