Twee bloke, A, met 'n massa van 4 kg, en B, met 'n massa van 9 kg, word met 'n ligte, onrekbare toutjie verbind - NSC Physical Sciences - Question 2 - 2023 - Paper 1

In the free-body diagram for block A, the forces acting on it include:

1. The weight of block A ($W_A = mg = 4kg imes 9.81m/s^2 = 39.24N$) acting downward.
2. The normal force ($N$) acting perpendicular to the inclined surface.
3. The tension ($T$) in the string acting upward along the incline.
4. The frictional force ($F_f$) acting opposite to the direction of motion along the incline.

Label these forces appropriately and ensure the angles are marked correctly.

To calculate the tension in the string, we must consider the forces acting on block A. Using the equation of motion along the incline:

$T - F_f - W_A imes ext{sin}( heta) = ma$ Substituting the values and solving for $T$, we find: $T = ma + F_f + W_A imes ext{sin}(35°)$ Insert the known values and solve.

The net force $F$ acting on block B can be calculated as follows:

$F = T + F_f + W_B imes ext{sin}(35°)$ Where:

• $T$ is the tension obtained in previous calculations,
• $F_f$ is the frictional force for block B, which can be similarly calculated,
• $W_B$ is the weight of block B ($9kg imes 9.81m/s^2$).

Substituting these values into the equation will yield the magnitude of the force $F$.

As die hoek van die skuinsvlak toeneem, sal die normale krag afneem, omdat die normale krag $N$ verband hou met die gewig van die blok wat op die skuinsvlak werk. Dit beteken dat die wrywingskrag, wat direk afhang van die normale krag ($f_K = ext{μ}N$), ook sal afneem. Dit kan lei tot 'n groter versnelling van die stelsels, aangesien daar minder weerstand teen die beweging is.