Power Output Calculation of a Motor

What is the power output of a motor with 40 oz-ft torque and 700-rpm speed?

A. 1/2 hp

B. 1/4 hp

C. 1/3 hp

D. 1/10 hp

Answer:

The motor will produce an output power of approximately 1/2 hp (A).

To determine the power output of a motor, we can use the formula:

Power (in horsepower, hp) = (Torque x Speed) / 5252

Given that the torque is 40 oz-ft and the speed is 700 rpm, we can calculate the power:

Power (in hp) = (40 oz-ft x 700 rpm) / 5252

Power (in hp) = 28000 / 5252 ≈ 5.32 hp

Next, we convert 5.32 hp to a fraction: 5.32 hp = 532/100 hp = 133/25 hp

Among the answer choices given, the closest fraction to 133/25 hp is 1/2 hp, making it the correct answer.

Therefore, the power output of the motor with 40 oz-ft torque and 700-rpm speed is approximately 1/2 hp.