A descending object has constant weight 40 N. Its upward air resistance is initially 8 N, later 40 N, and immediately after a parachute opens 80 N. It is moving downward at all three instants. (a) Determine the resultant force at each instant. (b) Describe how its speed changes. (c) Explain why terminal speed is not a state of zero force on the object.

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32 N downward, zero, 40 N upward; speed increases, is momentarily constant, then decreases. Choose downward as positive and subtract upward drag from weight. Initially F = 40 − 8 = 32 N downward, so the object accelerates in its direction of motion and gains speed. When drag is 40 N, the two forces balance and the resultant is zero. If this balance persists, acceleration is zero and the object continues downward at terminal speed. Just after the parachute opens, F = 40 − 80 = −40 N. This upward resultant opposes the downward velocity, so the object slows rather than immediately travelling upward. Terminal motion still involves both weight and air resistance; it is their vector sum that vanishes. As the open parachute slows, drag falls until a new balance can be reached at a smaller downward speed.

Forces during a descent 3

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