1. ## Physics Problem

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 582 N. As the elevator later stops, the scale reading is 399 N. Assume the magnitude of the acceleration is the same during starting and stopping.

1. Determine the weight of the person.
2. Determine the person's mass.
3. Determine the acceleration of the elevator.

2. Originally Posted by c_323_h
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 582 N. As the elevator later stops, the scale reading is 399 N. Assume the magnitude of the acceleration is the same during starting and stopping.

1. Determine the weight of the person.
2. Determine the person's mass.
3. Determine the acceleration of the elevator.
The Force on the scale is given by,
F=ma
The regular accelration is 9.8 m/s/s
Thus, when another a m/s/s is added we have,
582=m(9.8+a)
When it stops we have,
399=9.8m
Thus,
m=40.7 kilograms (am I right?)
Thus,
582=40.7(9.8+a)
We have,
a=4.49 m/s/s

Wow, I actually did a physics problem!
(Though I did not understand anything I was doing).

3. i thought mass was 40.7 too but it's wrong

4. Originally Posted by c_323_h
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 582 N. As the elevator later stops, the scale reading is 399 N. Assume the magnitude of the acceleration is the same during starting and stopping.

1. Determine the weight of the person.
2. Determine the person's mass.
3. Determine the acceleration of the elevator.
While the elevator is accelerating the weight (force) is:

M(-g+a)=582 N

While the elevator is decelleration the weight (force) is:

M(-g-a)=399 N.

where the g~=-9.91 m/s^2, and a is positive upwards.

-2Mg=981

or M= 50 kg

Which is the answer to part (2).

Now the weight is -gM=490.5 N; which is the answer to part (1)

Now we have:

M(-g+a)=582,

so:

490.5+50a=582,

so

a=1.83 m/s^2 (upwards)

RonL