# Chapter 4 - Dynamics: Newton's Laws of Motion - Problems: 17

a. $7.35 \frac{m}{s^{2}}$ in the downward direction. b. 1290 N.

#### Work Step by Step

a. The upward force of air resistance on the skydivers is (0.25)(mg). Apply Newton’s second law to the skydivers. Choose up to be the positive direction. $$\Sigma F=F_{air} – mg = ma$$ $$a = \frac{F_{air}-mg}{m} =\frac{0.25mg-mg}{m} = -0.75g \approx -7.35 \frac{m}{s^{2}}$$ The answer is rounded to 3 significant figures. The negative sign confirms that the acceleration is downward. The weight exceeds the air resistance force, so the skydivers’ acceleration is downward. b. If they are not accelerating, the magnitude of the upward air resistance force must equal their weight mg, so the net force is zero. $$F_{air} = 1.29 \times 10^{3} N$$ The answer is rounded to 3 significant figures.

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