A suitcase (mass m = 15 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.00 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Question

A suitcase (mass m = 15 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.00 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Answer 1

the force on the floor is m(g+a)

pressure is force/area

Answer 2

A suitcase of mass 16 kg

Answer 3

To find the pressure applied to the floor beneath the suitcase, we can use the formula:

Pressure = Force/Area

First, let's calculate the force exerted by the suitcase on the elevator floor.

The force exerted by the suitcase can be found using Newton's second law: F = ma

Since the elevator is accelerating upwards, the net force on the suitcase will be the difference between its weight and the force required to accelerate it upwards.

The weight of the suitcase is given by the equation: W = mg

So, the force exerted by the weight of the suitcase is F_weight = mg

The force required to accelerate the suitcase upwards is given by the equation: F_upwards = ma

Substituting the mass (m) and acceleration (a) values, we get F_upwards = 15 kg × 1.00 m/s² = 15 N

The net force on the suitcase can be calculated: F_net = F_upwards - F_weight

Substituting the values, we have F_net = 15 N - (15 kg × 9.8 m/s²) = 15 N - 147 N = -132 N

Note that the net force is negative because it opposes the weight of the suitcase.

Now, let's calculate the contact area between the suitcase and the floor.

The contact area is given as 0.50 m × 0.15 m = 0.075 m²

Finally, we can calculate the pressure using the formula:

Pressure = Force/Area

Substituting the values, we get Pressure = (-132 N) / (0.075 m²) = -1760 Pa

However, the question asks for the pressure in excess of atmospheric pressure. Atmospheric pressure at sea level is around 101,325 Pa.

Therefore, the pressure in excess of atmospheric pressure is:

Pressure_excess = Pressure - Atmospheric pressure

Substituting the values, we get Pressure_excess = -1760 Pa - 101,325 Pa = -103,085 Pa

Since pressure cannot be negative, we can express it as positive magnitude:

Pressure_excess = 103,085 Pa

So, the pressure applied to the floor beneath the suitcase, in excess of atmospheric pressure, is 103,085 Pa.

Answer 4

A suitcase (mass m = 15 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.40 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?