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Problem 15 Easy Difficulty

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 $\mathrm{N}$ and the drag force has a magnitude of 1027 $\mathrm{N}$ . The mass of the sky diver is 93.4 $\mathrm{kg}$ . What are the magnitude and direction of his acceleration?

Answer

$+1.20 \mathrm{m} / \mathrm{s}^{2}$
upward

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Video Transcript

we begin this question by calculating the Net force that acts on that's kind Ivor. So we have two forces that act on the vertical axis one apart and one downward. Then the net force. Is it close to the upward force minus the downward force? And this he's equals troop 112 do toes. So this is the net force acting on the skydiver. Now. In order to calculate his acceleration, we have to use Newton's second law, which says that the net force is equal to the mass times acceleration. Then we have to follow 112 Izzy close to 93.4 times acceleration. Then the acceleration is equals. True, 112 divided by 93.4, which is approximately 1.2 meters per second squared. So this is the magnitude off acceleration and note. The following the upward force is bigger than the downward force, So acceleration real point A ports