A man faces difficulty while swimming in a dam - NSC Physical Sciences - Question 2 - 2022 - Paper 1
Question 2
A man faces difficulty while swimming in a dam. During the rescue operation, an inflated tube attached to a helicopter by a rope is dropped from the helicopter.
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Worked Solution & Example Answer:A man faces difficulty while swimming in a dam - NSC Physical Sciences - Question 2 - 2022 - Paper 1
Step 1
2.1 State Newton’s First Law of Motion in words.
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Answer
Newton’s First Law of Motion states that a body will remain in its state of rest or uniform motion in a straight line unless acted upon by a resultant (non-zero) external force.
Step 2
2.2 Draw a free-body diagram of the man-tube combination while they are being dragged.
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The free-body diagram should include the following forces:
The gravitational force (
Fg=mg
) acting downwards on the man-tube combination.
The normal force (N) acting upwards from the water's surface.
The tension (T) in the rope at an angle of 50° with the horizontal.
The frictional force (f) acting horizontally against the direction of motion.
Step 3
2.3 Calculate the tension in the rope.
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To find the tension in the rope, we can use the following equation based on the forces acting on the man-tube combination:
ightarrow T ext{ is calculated as approximately } 466.79 ext{ N}.$$
Step 4
2.4 How will the answer to QUESTION 2.3 change if the helicopter ACCELERATES while dragging the man?
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If the helicopter accelerates, the net force on the man-tube combination will increase because an additional force will be needed to account for the acceleration. Hence, the tension in the rope will INCREASE. The frictional force remains constant, but the additional force needed to accelerate will result in a higher tension.
Step 5
2.5 Calculate the magnitude of the average upward force exerted on the inflated tube while it is sinking.
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Answer
To calculate the average upward force,
Calculate the gravitational force acting on the tube:
Fg=mimesg=4imes9.8=39.2extN
Determine the net force while sinking:
The displacement (s) is 0.8 m, and the initial speed (v_i) is 16 m/s. We can use:
vf2=vi2+2as
Assume the final speed (v_f) as 0 in water:
ightarrow a = -160 ext{ m/s}^2$$
Using Newton's second law:
Fnet=mimesaFnet=4imes(−160)=−640extN
The average upward force exerted (F_upward) can be calculated as:
Fupward=Fg+∣Fnet∣=39.2+640=679.2extN