A man faces difficulty while swimming in a dam - NSC Physical Sciences - Question 2 - 2022 - Paper 1

The free-body diagram consists of the following forces:

1. Weight (W) acting downwards due to gravity.
2. Normal force (N) acting perpendicular to the surface of the water.
3. Tension (T) in the rope acting at an angle of 50° to the horizontal.
4. Frictional force (f) acting opposite to the direction of motion (horizontal).

These forces should be labeled clearly on the diagram.

To calculate the tension in the rope, we use the equilibrium of forces in the vertical direction:

egin{align*} ext{Net force, } F_{net} &= 0 \ T ext{cos}(50°) - f - mg &= 0 \ T ext{cos}(50°) = f + mg \ T = \frac{f + mg}{ ext{cos}(50°)} ext{Substituting known values:} \ f = 300 ext{ N}, m = 70 ext{ kg}, g = 9.81 ext{ m/s}^2 \ T = \frac{300 + (70)(9.81)}{ ext{cos}(50°)} \ T \approx 468.75 ext{ N}. ext{Thus, the tension in the rope is approximately 468.75 N.} \end{align*}

If the helicopter accelerates while dragging the man, the net force acting on the man-tube combination increases. This results in an increase in the tension in the rope. Thus, the correct answer is: INCREASES.

To calculate the average upward force while the tube is sinking, we first use the kinematic equations:

Let: v_i = 16 ext{ m/s} \ (initial\ speed)
v_f = 0 \ (final\ speed\ when\ at\ the\ bottom)
\Delta y = -0.8 \text{ m} \ \text{(downwards)}
a = -160 ext{ m/s}^2

Using the formula: v_f^2 = v_i^2 + 2a\Delta y, egin{align*} 0 &= 16^2 + 2(-160)(-0.8) \ 0 = 256 + 256 \ ext{Net force, } F_{net} &= ma \ 679.20 ext{ N} = m(g + a)\ ext{(assuming } g = 9.81 ext{ m/s}^2)\ \F_{upward} = 679.20 ext{ N}. ext{Therefore, the magnitude of the average upward force exerted on the inflated tube is approximately 679.20 N.} \end{align*}