'n 20 kg-blok, wat op 'n ruwe horizontale oppervlak rus, is aan blokke P en Q verbind met 'n ligte onbreekbare tou/tjie wat oor 'n wrywingslose katrol beweeg - NSC Physical Sciences - Question 2 - 2020 - Paper 1

The free body diagram should include the following forces:

• Gravitational Force ( $F_g = mg = 20 ext{ kg} imes 9.81 ext{ m/s}^2 = 196.2 ext{ N}$).
• Normal Force (N) acting upward.
• Frictional Force (f) of 5 N acting opposite to the direction of motion.
• The applied force (F_a) of 35 N at an angle of 40° to the horizontal, which can be resolved into horizontal and vertical components.

Starting with Newton's second law, we have:

$F_{net} = ma$

Where the net force ($F_{net}$) is given by the equation:

$F_{net} = F_a - F_f$

Substituting the known forces:

$F_{net} = 35 ext{ N} - 5 ext{ N} = 30 ext{ N}$

Thus, we can rearrange the equation to find the mass:

$m = \frac{F_{net}}{g} = \frac{30 ext{ N}}{9.81 ext{ m/s}^2} ext{ kg} \\ m = 3.06 ext{ kg} \\ (approx.)$

As the block Q breaks and falls, the tension in the string will decrease because the force due to gravity will no longer be counteracted by the tension from the suspended block.

Once block Q falls, the acceleration of the remaining block to the left will increase due to the net force acting on it without the counteracting force from block Q.