Question

A suitcase (mass m = 18 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 1.6 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Answer 1

To calculate the pressure applied to the floor beneath the suitcase, multiply the mass of the suitcase by the acceleration of the elevator and divide by the area of contact.

Step-by-step explanation:

To calculate the pressure applied to the floor beneath the suitcase, we need to use the equation:

Pressure = Force / Area

The force can be found by using Newton's second law:

Force = mass x acceleration

Substituting the given values, we get:

Force = 18 kg x 1.6 m/s^2 = 28.8 N

The area is given as 0.50 m x 0.15 m = 0.075 m^2

Now we can calculate the pressure:

Pressure = 28.8 N / 0.075 m^2 = 384 Pa

Answer 2

P=2736 Pa

Step-by-step explanation:

According to Newton we have that:

∑
`F=m*a\\`

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

∑
`F=m*g+m*a=m*(g+a)`

`F=205.2N`

We know that the pressure is given by:

`P=(F)/(A)\\P=(205.2N)/((0.50m)*(0.15m))\\P=2736N/m^2=2736Pa`