Question

the upward normal force exerted by the floor is 620 n on an elevator passenger who weighs 650 n. you may want to review (page). for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of determining acceleration from force. what is the magnitude of the acceleration? express your answer in meters per second squared. view available hint(s) a = ____ m/s²

Explanation:

Step1: Find mass of passenger

Weight \( W = mg \), so mass \( m=\frac{W}{g} \). Given \( W = 650\ N \), \( g = 9.8\ m/s^2 \), so \( m=\frac{650}{9.8}\ kg \approx 66.33\ kg \).

Step2: Apply Newton's second law

Forces on passenger: normal force \( N = 620\ N \) (upward), weight \( W = 650\ N \) (downward). Net force \( F_{net}=W - N \) (downward, so acceleration is downward). By \( F_{net}=ma \), \( a=\frac{W - N}{m} \). Substitute \( W = 650\ N \), \( N = 620\ N \), \( m\approx66.33\ kg \): \( a=\frac{650 - 620}{66.33}=\frac{30}{66.33}\approx - 0.452\ m/s^2 \) (negative sign means downward acceleration, magnitude is \( 0.452\ m/s^2 \)).

Answer:

\( 0.45\) (or more precise \( 0.452\)) \( m/s^2 \) (magnitude, so we can write \( 0.45\ m/s^2 \) or \( 0.452\ m/s^2 \))