Question

A person stands on a scale in an elevator. The maximum and minimum scale readings are 814.7 N and 321.8 N, respectively. The acceleration of gravity is 9.8 m/s 2 . Assume the magnitude of the acceleration is the same during starting and stopping, and determine the weight of the person. Answer in units of N.

Answer 1

Step-by-step explanation:

Maximum scale reading, R = 814.7 N

Minimum scale reading, R' = 321.8 N

acceleration due to gravity, g = 9.8 m/s²

let a be the acceleration of the elevator.

Case 1: Elevator is moving up :

R - mg = ma

814.7 - mg = ma

814.7 = m (g + a) .... (1)

Case 2: elevator is moving down :

mg - R' = ma

R' = m (g - a)

321 = m (g - a) .... (2)

divide equation (1) by (2)

2.53 (g - a) = g + a

25 - 2.53 a = 9.8 + a

15.2 = 3.53 a

a = 4.31 m/s²

Put in equation (1)

814.7 = m (9.8 + 4.31)

m = 57.8 kg

Answer 2

the weight of the person is 57.98kg

Step-by-step explanation:

According to Newton's second law:

`\sum F=m*a`

The scale reads the normal force, so if the elevator is going upward:

`N-m.g=m.a`

`N-m*(9.8)=m.a`

and if it is going downward:

`N-m.g=m.a`

`N-m*(9.8)=-m.a`

so:

`814.7N-m*(9.8)=m.a`

`321.8N-m*(9.8)=-m.a`

because the value of the acceleration is equal in magnitude we can substitute one equation into the other.

`814.7N-m*(9.8)=m.((m*9.8-321.8N)/(m))\\\\m=(814.7+321.8)/(2*9.8)\\\\m=57.98kg`