A 76.6-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 1.94 m/s2, (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.15 m/s2?

I've already figured out part (b) which is just 750.68 N, I need help with parts (a) and (c)

Just add/subtract the given acceleration to gravity. Then as usual, F=ma, where a is the net acceleration.

To solve parts (a) and (c), we need to consider the forces acting on the person in each situation. Let's go step-by-step to find the apparent weight in each case:

(a) Accelerating upward with an acceleration of 1.94 m/s^2:
When the elevator is accelerating upward, there are two forces acting on the person: their weight (mg) pointing downward, and the normal force (N) exerted by the scale pointing upward. The normal force is what the scale measures as the person's weight.

1. Calculate the net force:
The net force acting on the person is given by the equation: F_net = m * a
where m is the mass of the person and a is the acceleration of the elevator.
F_net = (76.6 kg) * (1.94 m/s^2)
F_net = 148.604 N

2. Calculate the apparent weight:
To find the apparent weight, we need to take into account both the person's weight and the net force acting on them. The apparent weight is equal to the sum of the person's weight and the normal force.
Apparent weight = weight + net force
Apparent weight = mg + F_net
Apparent weight = (76.6 kg) * (9.8 m/s^2) + 148.604 N
Apparent weight = 750.68 N + 148.604 N
Apparent weight = 899.284 N

Therefore, the apparent weight of the person when the elevator is accelerating upward with an acceleration of 1.94 m/s^2 is 899.284 N.

(c) Accelerating downward with an acceleration of 1.15 m/s^2:
Following the same steps as in part (a):

1. Calculate the net force:
The net force acting on the person is given by the equation: F_net = m * a
F_net = (76.6 kg) * (-1.15 m/s^2) [note the negative sign for downward acceleration]
F_net = -88.19 N

2. Calculate the apparent weight:
The apparent weight is again equal to the sum of the weight and the net force.
Apparent weight = weight + net force
Apparent weight = mg + F_net
Apparent weight = (76.6 kg) * (9.8 m/s^2) - 88.19 N
Apparent weight = 750.68 N - 88.19 N
Apparent weight = 662.49 N

Therefore, the apparent weight of the person when the elevator is accelerating downward with an acceleration of 1.15 m/s^2 is 662.49 N.

To solve parts (a) and (c), we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the net force acting on the person is equal to the gravitational force (weight) minus the force exerted by the elevator floor.

(a) When the elevator is accelerating upward with an acceleration of 1.94 m/s^2:
To find the apparent weight, we need to determine the net force acting on the person. The force exerted by the floor is equal to the person's mass multiplied by the acceleration of the elevator.
The formula to calculate the apparent weight is:
Apparent weight = Weight - Force exerted by the floor

Weight = mass * gravity
= 76.6 kg * 9.8 m/s^2 (acceleration due to gravity)
= 750.28 N

Force exerted by the floor = mass * acceleration of the elevator
= 76.6 kg * 1.94 m/s^2
= 148.604 N

Apparent weight = 750.28 N - 148.604 N
= 601.676 N

Therefore, the apparent weight when the elevator is accelerating upward with an acceleration of 1.94 m/s^2 is 601.676 N.

(c) When the elevator is accelerating downward with an acceleration of 1.15 m/s^2:
The steps to find the apparent weight in this case are the same as in part (a), except the acceleration is negative since it is downward.
Force exerted by the floor = mass * acceleration of the elevator
= 76.6 kg * (-1.15 m/s^2)
= -88.39 N

Apparent weight = 750.28 N - (-88.39 N)
= 838.67 N

Therefore, the apparent weight when the elevator is accelerating downward with an acceleration of 1.15 m/s^2 is 838.67 N.