A 20 kg block, resting on a rough horizontal surface, is connected to blocks P and Q by a light inextensible string moving over a frictionless pulley - NSC Physical Sciences - Question 2 - 2020 - Paper 1

The free-body diagram of the 20 kg block includes the following forces:

1. The gravitational force acting downwards, represented as $F_g = mg = 20 ext{ kg} imes 9.8 ext{ m/s}^2 = 196 ext{ N}$.
2. The normal force acting upwards, denoted as $F_N$.
3. The applied force of 35 N at an angle of 40°, which can be broken into components:
   • Horizontal component: $F_{ax} = 35 ext{ N} imes ext{cos}(40°)$
   • Vertical component: $F_{ay} = 35 ext{ N} imes ext{sin}(40°)$
4. The frictional force acting against the direction of motion, $f = 5 ext{ N}$.

From the equilibrium of forces in the horizontal direction with constant speed:

Net force $F_{net} = 0$, hence:

$F_{applied} - f - F_{ax} = 0$

Substituting the values:

$35 ext{ N} - 5 ext{ N} - (35 ext{ N} imes ext{cos}(40°)) = 0$

Solving for combined mass $m$ gives:

Since the tension in the system is balanced and the system is at constant speed:
m = rac{5 N}{g} + rac{F_{applied}}{g} = rac{5 N + 35 ext{ N} imes ext{sin}(40°)}{9.8 ext{ m/s}^2} = 3.25 ext{ kg}.

Decreases. When block Q breaks off, the tension in the string is no longer supporting the weight of block Q, leading to a decrease in the overall system tension.

Decreases. If block Q is no longer attached, the system dynamics change, and the remaining force results in a net backward force, thus reducing the velocity of the 20 kg block.