Question

Determine the work done by an electric motor lifting an elevator and its occupants weighing 2200lb a distance of 60ft. Compute the power requirements for the motor complete the lift in: a. 5sec[W=132,000lb⋅ft,P=48HP] b. 8sec[W=132,000lb⋅ft,P=30HP] Express your results in horsepower ( 1hp=550ftlb/s).

Answer 1

The work done by an electric motor to lift an elevator weighing 2200 pounds a height of 60 feet is 132,000 lb·ft. The power requirements are 48 HP for a 5-second lift and 30 HP for an 8-second lift.

Step-by-step explanation:

The work done by an electric motor in lifting an elevator can be calculated using the formula W = weight × height.

First, we convert the weight from pounds to pound-feet by multiplying by the height:

• W = 2200lb × 60ft = 132,000lb·ft

Next, we calculate the power requirement for the motor using the formula P = W / t, where 't' is time in seconds. Since 1 horsepower (HP) is equal to 550 ft·lb/s, we can convert our power result to horsepower.

1. For a 5-second lift: P = 132,000lb·ft / 5s = 26,400 ft·lb/s. Then, convert to horsepower: 26,400 ft·lb/s / 550 ft·lb/s per HP = 48 HP.
2. For an 8-second lift: P = 132,000lb·ft / 8s = 16,500 ft·lb/s. Convert to horsepower: 16,500 ft·lb/s / 550 ft·lb/s per HP = 30 HP.

Therefore, the power requirements for the motor to complete the lift are 48 HP for a 5-second lift and 30 HP for an 8-second lift.