Block P, of mass 2 kg, is connected to block Q, of mass 3 kg, by a light inextensible string - NSC Physical Sciences - Question 2 - 2019 - Paper 1

In the free-body diagram for block P, include the following forces:

1. Gravitational force (Weight) acting downwards: $F_g = m imes g$ where $m = 2 ext{ kg}$ and $g ext{ (acceleration due to gravity)} ext{ is approximately } 9.81 ext{ m/s}^2$.
2. Normal force acting perpendicular to the incline.
3. Frictional force acting opposite to the direction of motion (2.5 N).
4. Tension force in the string pulling block P towards block Q.

The forces acting on block P can be examined using the formula:

$F_{net} = T - F_f - F_{gx}$

Where:

• $T$ is the tension in the string
• $F_f = 2.5 ext{ N}$ (frictional force)
• $F_{gx} = mg ext{sin}( heta)$ where $heta = 30°$
• For block P: $F_{gx} = (2 ext{ kg})(9.81 ext{ m/s}^2)( ext{sin}(30°)) = 9.81 ext{ N} imes 0.5 = 4.905 ext{ N}$.

Thus, we can rearrange the forces: $F_{net} = T - F_f - F_{gx}$ Substituting, we calculate: $F_{net} = T - 2.5 ext{ N} - 4.905 ext{ N}$ Assuming the tension $T$ due to block Q’s weight and force: When solving, if we assume the tension is equal to the net force acting down the incline, we can calculate the acceleration as: a = rac{F_{net}}{m} = rac{T - 2.5 - 4.905}{2} = rac{T - 7.405}{2}.

Further calculations might be required based on the tension provided by block Q.

The acceleration of block P after passing point B will be SMALLER THAN the acceleration calculated in QUESTION 2.3. This is because, after point B, the frictional force is no longer acting against it due to the incline being smooth, which generally allows for an increase in potential energy causing a decrease in net force acting down the incline.