Question

A 60.0-kg person rides in an elevator while standing on a scale. The elevator is traveling downward but slowing down at a rate of 2.00 m/s². The reading on the scale is closest to:

Option 1: 708 N
Option 2: 469 N
Option 3: 120 N
Option 4: 589 N

Answer 1

The reading on the scale would be closest to 469 N, as the 60-kg person is in a decelerating elevator, which decreases the normal force compared to the person's actual weight.

Step-by-step explanation:

The question concerns a 60.0-kg person standing on a scale in a descending elevator which is decelerating at 2.00 m/s². To find the reading on the scale, we must consider the apparent weight, which is the normal force the scale exerts to support the person. The real weight of the person (W = mg) would be 60.0 kg × 9.81 m/s² = 588.6 N. Since the elevator is decelerating, the net acceleration will be g - a, thus:

Net force (apparent weight) = mass × (g - a).

For this case: Net force = 60.0 kg × (9.81 m/s² - 2.00 m/s²) = 60.0 kg × 7.81 m/s² = 468.6 N.

The reading on the scale (closest to 469 N) reflects the decreased normal force due to the downward acceleration subtracted from gravity.