Question

John weight's on the ground is 145lb. John enters an elevator and stands on a scale. In its way up, the elevator is able to uniformly accelerate from zero to 2m/s in 5s.

a. What does the scale read before the elevator starts moving up
b. While the elevator is accelerating up, what is John's weight
c. After 10s, the elevator continues going up at constant speed, what does the scale reads now?
d. After getting to the top floor, the elevator starts moving down with an acceleration of 1.3m/s2, what is the read in the scale then?

Answer 1

John's apparent weight varies due to the acceleration of the elevator. It increases when the elevator accelerates upward, returns to his actual weight when moving at constant speed, and decreases when the elevator accelerates downward.

Step-by-step explanation:

John's weight on the scale before the elevator starts moving up is equivalent to his actual weight on the ground, 145lb, which converts to 644.1 N (since weight is a force measured in Newtons and 1 lb = 4.44822 N).

When the elevator is accelerating upward, John's apparent weight is the sum of his actual weight and the force of acceleration. To find the scale reading while the elevator is accelerating up, use the formula Fs = ma + mg. John's mass (m) is 65.77 kg (145lb / 2.20462 lb/kg). The acceleration (a) is 2 m/s², and g is the acceleration due to gravity (9.8 m/s²). The scale reading (Fs) is thus:

Fs = 65.77 kg * 2 m/s² + 65.77 kg * 9.8 m/s² = 776.7 N.

After 10 seconds, when the elevator is moving at constant speed, the acceleration is zero, hence there's no additional force exerted by the elevator, and the reading returns to John's actual weight, 644.1 N.

While the elevator is accelerating downward at 1.3 m/s², the scale reads less than John's actual weight. Using Fs = m(g - a), the scale reading would be:

Fs = 65.77 kg * (9.8 m/s² - 1.3 m/s²) = 560.7 N.