Forces and Motion in Elevators

When a person is standing in an elevator accelerating upward, the normal force from the floor is ________ compared to when the elevator is at rest.

• greater than

• equal to

• less than

• zero

Answer explanation

When the elevator accelerates upward, the person experiences an additional upward force due to acceleration. This increases the normal force from the floor, making it greater than when the elevator is at rest.

Using Newton's Second Law, if an elevator accelerates upward at 2 m/s², what is the normal force on a 70 kg passenger?

• 686 N

• 826 N

• 588 N

• 700 N

Answer explanation

Using F = ma, the weight (mg) is 70 kg * 9.8 m/s² = 686 N. The net force is F_net = ma = 70 kg * 2 m/s² = 140 N. The normal force (N) is N = weight + F_net = 686 N + 140 N = 826 N. Thus, N = 784 N.

When an elevator is in free fall (ignoring air resistance), the normal force on a passenger is:

• equal to their weight

• greater than their weight

• less than their weight

• zero

Answer explanation

When an elevator is in free fall, both the elevator and the passenger accelerate downwards at the same rate due to gravity. Therefore, the normal force acting on the passenger is zero, as there is no contact force opposing their weight.

What causes the 'weightless' feeling when an elevator initially starts moving down?

• Decrease in normal force

• Increase in gravitational force

• Increase in air pressure

• Decrease in mass

Answer explanation

The 'weightless' feeling occurs because the normal force acting on a person decreases as the elevator accelerates downward. This reduction in normal force makes it feel as if one is lighter, creating the sensation of weightlessness.

If F = ma, and a person's apparent weight changes in an elevator, what remains constant?

• The normal force

• The person's mass

• The acceleration due to gravity

• The net force

Answer explanation

In the equation F = ma, the person's mass remains constant regardless of the elevator's motion. While the normal force and apparent weight change, mass does not, as it is an intrinsic property of the person.

When an elevator decelerates while moving upward, the normal force on a passenger is ________ than when at rest.

• greater

• less

• equal

• zero

Answer explanation

When the elevator decelerates while moving upward, the effective acceleration acting on the passenger decreases. This results in a lower normal force compared to when at rest, making the normal force less.

Calculate the apparent weight of a 50 kg person in an elevator accelerating upward at 3 m/s². (g = 9.8 m/s²)

• 490 N

• 640 N

• 590 N

• 540 N

Answer explanation

To find the apparent weight, use the formula: Apparent Weight = m(g + a). Here, m = 50 kg, g = 9.8 m/s², and a = 3 m/s². Thus, Apparent Weight = 50(9.8 + 3) = 50(12.8) = 640 N. The correct answer is 640 N.

The difference between actual weight and apparent weight in an elevator is directly proportional to:

• The elevator's velocity

• The elevator's mass

• The elevator's acceleration

• The elevator's height

Answer explanation

The difference between actual weight and apparent weight in an elevator is influenced by the elevator's acceleration. When the elevator accelerates, it affects the forces acting on the person inside, altering their apparent weight.

In an elevator moving at constant velocity, the normal force is:

• Equal to the weight

• Greater than the weight

• Less than the weight

• Zero

Answer explanation

In an elevator moving at constant velocity, the forces are balanced. The normal force acting on the elevator is equal to the weight of the object inside, as there is no acceleration to change this balance.

If a person experiences an apparent weight of 1.5 times their normal weight in an elevator, what is the elevator's acceleration? (g = 9.8 m/s²)

• 4.9 m/s² upward

• 9.8 m/s² upward

• 4.9 m/s² downward

• 9.8 m/s² downward

Answer explanation

The apparent weight is 1.5 times the normal weight, meaning the net force is 1.5mg. Using F = ma, we find a = (1.5mg - mg)/m = 0.5g. Thus, a = 0.5 * 9.8 m/s² = 4.9 m/s² upward.