# Chapter 4 - Dynamics: Newton's Laws of Motion - Problems: 16

$2.5 \frac{m}{s^{2}}$ in the downward direction.

#### Work Step by Step

The upward normal force on the woman from the scale is (0.75)(mg). Apply Newton’s second law to the woman. Choose up to be the positive direction. $$\Sigma F=F_n – mg = ma$$ $$a = \frac{F_n-mg}{m} =\frac{0.75mg-mg}{m} = -0.25g \approx -2.5 \frac{m}{s^{2}}$$ The answer is rounded to 2 significant figures. The negative sign confirms that the acceleration is downward. The weight exceeds the normal force, so the woman’s acceleration is downward. The initially motionless elevator must have begun to move downward.

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