Question

A person stands on a scale in an elevator. As the elevator starts,

the scale has a constant reading of 591 N. As the elevator later

stops, the scale reading is 391 N. Assume the magnitude of the

acceleration is the same during starting and stopping, and

determine (a) the weight of the person, (b) the person’s mass, and

(c) the acceleration of the elevator
. proper matter​

Answer 1

Answer:

answer Is

Explanation:

The scale reads the upward normal force exerted by the floor on the passenger. The maximum force occurs during upward acceleration (when starting an upward trip or ending a downward trip). The

minimum normal force occurs with downward acceleration. For each respective situation,

∑F

y

=ma

y

becomes for starting +591N−mg=+ma

and for stopping +391N−mg=−ma

where a represents the magnitude of the acceleration.

(a) These two simultaneous equations can be added to eliminate a and solve for mg:

+591N−mg+391N−mg=0

or 982N–2mg=0

F

g

=mg=

2

982N

=491N

(b) From the definition of weight, m=

g

F

g

=

9.80m/s

2

491N

=50.1kg

(c) Substituting back gives +591N−491N=(50.1kg)a, or

a=

50.1kg

100N

=2.00m/s

2

Was this answer helpful?