Question

1. a student stands on a platform scale inside of an elevator. the student has a mass of 70 kg.

a. the student presses a button to go up from the first floor to the 20th floor.
i. initially the elevator must have an acceleration that points as it starts to move.
ii. draw a fbd for the forces acting on the student during this time.
iii. what type of force does a platform scale measure?
iv. during this time the platform scale has a reading of 1036 newtons. write a sum of the forces and use it to calculate the acceleration of the student inside the elevator.
b. after some time the elevator comes to a constant velocity as it moves upward.
i. what must be the acceleration of the elevator now?
ii. draw a fbd for the forces acting on the student during this time.
iii. what reading does the platform scale show now? show your work below.

Explanation:

Step1: Determine acceleration when velocity is constant

When an object moves with a constant velocity, its acceleration \(a = 0\) according to the formula \(a=\frac{v - u}{t}\), where \(v = u\) (final velocity equals initial velocity).

Step2: Draw FBD for constant - velocity case

The forces acting on the student are the normal force \(F_N\) (upward) and the gravitational force \(F_g=mg\) (downward). Since \(a = 0\), from Newton's second law \(F_{net}=ma = 0\), so \(F_N=F_g\). The FBD will have two vectors of equal length: \(F_N\uparrow\) and \(F_g\downarrow\).

Step3: Calculate scale reading

Using Newton's second law \(F_{net}=F_N - F_g=ma\). Since \(a = 0\), \(F_N=F_g\). Given \(m = 70\space kg\) and \(g = 9.8\space m/s^2\), \(F_g=mg=70\times9.8 = 686\space N\). So the scale reading (which measures \(F_N\)) is \(686\space N\).

Answer:

i. The acceleration of the elevator is \(0\space m/s^2\).
ii. The FBD has two vectors: normal force \(F_N\) (upward) and gravitational force \(F_g\) (downward) of equal length.
iii. The scale reading is \(686\space N\).