In this question use $g = 9.8 \, \text{ms}^{-2}$ - AQA - A-Level Maths: Mechanics - Question 16 - 2017 - Paper 2

In this step, you would need to:

1. Draw a diagram of the box on the inclined board at an angle of $5^{\circ}$.
2. Indicate the following forces:
   • Weight $W = mg$ acting downwards.
   • Normal force $R$ acting perpendicular to the inclined plane.
   • Tension $T$ acting along the string at an angle of $40^{\circ}$.
   • Frictional force $F_f = \mu R$ acting opposite to the direction of motion along the incline.

Ensure the diagram clearly labels each force with arrows indicating direction and maintains proper scaling where possible.

To find the tension in the string when the box is accelerating, we use Newton's second law:

1. Force along the incline:

   $F_{net} = ma$

   The net force acting along the incline where the box accelerates is:

   $T \cos(40^{\circ}) - mg \sin(5^{\circ}) - \mu R = ma$

2. Resolve the forces:

   We already have from part (b):

   • Normal force $R$ from previous calculations as $40.0984 \, \text{N}$.
   • The weight component parallel to the plane is $mg \sin(5^{\circ})$.

3. Setting up the equation:

   Substituting known values:

   $\Rightarrow T \cos(40^{\circ}) - 8.0 \times 9.8 \sin(5^{\circ}) - 0.83 \cdot 40.0984 = 8.0 \cdot 3$

   Simplifying this expression:

   Calculate $8.0 \times 9.8 \sin(5^{\circ})$ which is approximately: $8.0 \times 9.8 \times 0.0872 \approx 6.8514$

   Substitute:

   $T \cos(40^{\circ}) - 6.8514 - 33.3889 = 24$

   Thus:

   $T \cos(40^{\circ}) = 24 + 6.8514 + 33.3889 = 64.2403$

   Finally, solve for $T$:

   $T = \frac{64.2403}{\cos(40^{\circ})}$

   Compute the value:

   $T \approx 64.2403 / 0.7660 \approx 83.77 \, \text{N}.$