A student stands on a bathroom scale in an

Question

elevator at rest on the 64th floor of a building.
The scale reads 843 N. As the elevator moves
up, the scale reading increases to 936 N, then
decreases back to 843 N.
The acceleration of gravity is 9.8 m/s
2
.
Find the acceleration of the elevator.
Answer in units of m/s2

Answer 1

To find the acceleration of the elevator, we can use the readings from the bathroom scale.

1. Weight at rest: When the elevator is at rest, the scale reads 843 N. The weight (W) of the student can be computed using the equation: \[ W = m \cdot g \] where \( W = 843 , \text{N} \) and \( g = 9.8 , \text{m/s}^2 \).

   We can solve for mass \( m \): \[ m = \frac{W}{g} = \frac{843 , \text{N}}{9.8 , \text{m/s}^2} \approx 86.1 , \text{kg} \]

2. Reading when accelerating up: When the elevator accelerates upward, the scale reads 936 N. The effective weight \( W' \) when accelerating can be calculated using the equation: \[ W' = m \cdot (g + a) \] where \( a \) is the acceleration of the elevator. Setting this equal to the scale reading when accelerating: \[ 936 , \text{N} = 86.1 , \text{kg} \cdot (9.8 , \text{m/s}^2 + a) \]

3. Setting up the equation: \[ 936 = 86.1 \cdot (9.8 + a) \]

   Dividing both sides by 86.1: \[ \frac{936}{86.1} = 9.8 + a \]

   Calculating \( \frac{936}{86.1} \): \[ 10.85 \approx 9.8 + a \]

4. Solving for \( a \): \[ a = 10.85 - 9.8 = 1.05 , \text{m/s}^2 \]

Thus, the acceleration of the elevator when it is moving upward is approximately 1.05 m/s².