A person gets into an elevator at the lobby level of a hotel together with his 28-kg suitcase, and gets out at the 10th floor 35 m above. Determine the amount of energy consumed by the motor of the elevator that is now stored in the suitcase. Assume that the vibrational effects in the elevator are negligible.

P.E = 9604J = 9.60KJ

Therefore, the amount of energy consumed by the motor of the elevator that is now stored in the suitcase is 9604J

Explanation:

The amount of energy consumed by the motor of the elevator that is now stored in the suitcase is equal to the potential energy gained by the suitcase.

Potential energy is the energy by virtue of its position. And can be written as;

P.E = mgh

Given

mass m = 28kg

height h = 35m

Acceleration due to gravity g = 9.8m/s^2

P.E = 28 × 35 × 9.8

P.E = 9604J = 9.60KJ

Therefore, the amount of energy consumed by the motor of the elevator that is now stored in the suitcase is 9604J

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq