Question

1. an 784 n woman stands on a scale in a motionless elevator. when the elevator begins to move, the scale reads 600 n. what is the magnitude and direction of the elevator’s acceleration?

Explanation:

Step1: Find the mass of the woman

We know that weight \( W = mg \), where \( g = 9.8\ m/s^2 \). Given \( W = 784\ N \), we can find mass \( m \) as \( m=\frac{W}{g} \).
\( m = \frac{784\ N}{9.8\ m/s^2}=80\ kg \)

Step2: Analyze the forces on the woman

When in the elevator, the forces on the woman are the gravitational force (weight) \( W = mg \) downward and the normal force \( N \) (scale reading) upward. Using Newton's second law \( F_{net}=ma \), where \( F_{net}=W - N \) (since \( W \) is downward and \( N \) is upward, the net force is downward if \( W>N \)).
So \( F_{net}=784\ N - 600\ N = 184\ N \) (downward)

Step3: Calculate the acceleration

From \( F_{net}=ma \), we can solve for \( a \): \( a=\frac{F_{net}}{m} \)
Substituting \( F_{net}=184\ N \) and \( m = 80\ kg \), we get \( a=\frac{184\ N}{80\ kg}=2.3\ m/s^2 \)
The direction of acceleration is downward (since the net force is downward, and acceleration is in the direction of net force).

Answer:

The magnitude of the elevator's acceleration is \( 2.3\ m/s^2 \) and the direction is downward.