Question

diagnostic questions 5 - 2 a 130lb woman stands in an elevator on a scale as the elevator just begins to move upward from the first to second floor. which choice could properly describe her weight, contact force, and acceleration at this instant? your response history: 1. incorrect. your answer: a. weight = 130 lbs downward, contact force = 130 lbs upward, acceleration is zero 2. incorrect. your answer: d. weight = 140 lbs downward, contact force = 130 lbs upward, acceleration is downward score: 0/0.19 you may change your answer and resubmit: a. weight = 130 lbs downward, contact force = 130 lbs upward, acceleration is zero b. weight = 130 lbs downward, contact force = 130 lbs upward, acceleration is downward c. weight = 130 lbs downward, contact force = 130 lbs upward, acceleration is upward d. weight = 140 lbs downward, contact force = 130 lbs upward, acceleration is downward e. weight = 130 lbs downward, contact force = 140 lbs upward, acceleration is upward f. weight = 130 lbs downward, contact force = 140 lbs upward, acceleration is downward

Explanation:

Step1: Analyze forces acting on woman

The weight of the woman \(W = 130\) lbs acts downward due to gravity. It is a constant for this problem as the mass and gravitational - acceleration near the Earth's surface are constant.

Step2: Apply Newton's second law \(F_{net}=ma\)

When the elevator accelerates upward, the net force \(F_{net}\) acting on the woman is \(F_{net}=N - W\), where \(N\) is the contact force (normal force) exerted by the scale on the woman. According to Newton's second law \(F_{net}=ma\), and since \(a>0\) (upward acceleration), \(N - W>0\), so \(N>W\).

Step3: Determine correct option

Since \(W = 130\) lbs downward and \(N>W\) when accelerating upward, the contact force \(N = 140\) lbs upward and acceleration is upward.

Answer:

E. weight = 130 lbs downward, contact force = 140 lbs upward, acceleration is upward