An airplane is undergoing horizontal acceleration of 5m/s2. A block weighing 2kg is suspended from the airplane, initially at rest relative to the airplane. What is the angle θ that the string makes with the horizontal.

To determine the angle $\theta$ that the string makes with the horizontal when the airplane is undergoing horizontal acceleration, we need to analyze the forces acting on the block. Here are the steps to solve the problem:

1. Identify the forces acting on the block:

   • The weight of the block ($W$) acts vertically downward. This force is given by $W = mg$, where $m$ is the mass of the block and $g$ is the acceleration due to gravity.
   • The tension in the string ($T$) acts along the string.
   • The horizontal acceleration of the airplane ($a$) causes a horizontal force on the block.

2. Set up the coordinate system:

   • Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.

3. Resolve the forces into components:

   • The tension $T$ in the string can be resolved into two components:
     • $T_x$ (horizontal component) = $T \cos \theta$
     • $T_y$ (vertical component) = $T \sin \theta$
   • The weight $W$ acts vertically downward: $W = mg$.

4. Apply Newton's second law in the horizontal direction:

   • The horizontal force is provided by the tension component $T_x$: $T \cos \theta = ma$
   • Here, $a = 5 \, \text{m/s}^2$ and $m = 2 \, \text{kg}$.

5. Apply Newton's second law in the vertical direction:

   • The vertical forces must balance since there is no vertical acceleration: $T \sin \theta = mg$
   • Here, $g = 9.8 \, \text{m/s}^2$.

6. Solve the equations:

   • From the vertical force balance: $T \sin \theta = 2 \times 9.8 = 19.6 \, \text{N}$
   • From the horizontal force balance: $T \cos \theta = 2 \times 5 = 10 \, \text{N}$

7. Find the angle $\theta$:

   • Divide the horizontal force equation by the vertical force equation to eliminate $T$: $\frac{T \cos \theta}{T \sin \theta} = \frac{10}{19.6}$ $\cot \theta = \frac{10}{19.6}$ $\cot \theta = 0.5102$
   • Take the arctangent to find $\theta$: $\theta = \arctan \left( \frac{1}{0.5102} \right)$ $\theta = \arctan (1.96)$ $\theta \approx 63.4^\circ$

Therefore, the angle $\theta$ that the string makes with the horizontal is approximately $63.4^\circ$.