A 97.0-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is

(a) accelerating upward with an acceleration of 1.70 m/s2,

(b) moving upward at a constant speed, and

(c) accelerating downward with an acceleration of 1.80 m/s2?

a. Since W=mass x acceleration , when accelerating upward the acceleration due to gravity and the acceleration of the elevator add up. So, a=(9.8+1.7)=11.5m/s^2, so weight = 97kg x 11.5 m/s^2.

b. at constant speed their is no acceleration except for earths gravity which means its weight is the same as if its not moving.

c. if it accelerates downward use the same concept as a except subtract the acceleration from the acceleration due to gravity.

Thanks Spencer !

Thank you Spencer

To find the apparent weight of a person in different elevator scenarios, we need to consider the forces acting on the person.

(a) When the elevator is accelerating upward with an acceleration of 1.70 m/s^2:
The apparent weight of the person will be the normal force acting on them. In this scenario, the net force acting on the person is the difference between their weight and the force due to acceleration.
Net force = mass * acceleration
Net force = 97.0 kg * 1.70 m/s^2
Net force = 164.9 N

Since the person is in equilibrium (not moving up or down), the net force is equal to the normal force. Therefore, the apparent weight of the person is 164.9 N.

(b) When the elevator is moving upward at a constant speed:
In this scenario, the elevator is not accelerating, which means there is no net force acting on the person. The only force acting on them is their weight due to gravity. Therefore, the apparent weight of the person will be their actual weight, which is 97.0 kg * 9.8 m/s^2 = 950.6 N.

(c) When the elevator is accelerating downward with an acceleration of 1.80 m/s^2:
Similar to the first scenario, we need to calculate the net force acting on the person.
Net force = mass * acceleration
Net force = 97.0 kg * (-1.80 m/s^2) (Note the negative sign indicates the opposite direction of acceleration)
Net force = -174.6 N

Again, since the person is in equilibrium, the net force is equal to the normal force. However, in this case, the normal force is less than the person's weight due to the downward acceleration. Therefore, the apparent weight of the person is 174.6 N less than their actual weight. We need to subtract this value from the weight to find the apparent weight:
Apparent weight = 950.6 N - 174.6 N = 776 N.

To summarize, the apparent weight of the person in the different elevator scenarios are:
(a) 164.9 N
(b) 950.6 N
(c) 776 N