Question:

A 500 W motor is used to lift a 200 kg load vertically upward. The load is lifted a distance of 10 m in 5 s. Determine the efficiency of the motor in converting electrical power to mechanical power.

Answer:

To determine the efficiency of the motor, we first need to calculate the mechanical power output and the electrical power input.

Given: Electric power input (P_in) = 500 W Distance lifted (d) = 10 m Time taken (t) = 5 s Mass of the load (m) = 200 kg

To calculate the mechanical power output (P_out), we'll use the formula:

P_out = W / t,

where W is the work done, given by:

W = m * g * d,

where g is the acceleration due to gravity (9.8 m/s²).

Calculating W: W = (200 kg) * (9.8 m/s²) * (10 m) = 196,000 J

Plugging the values of W and t into the formula for P_out: P_out = W / t = 196,000 J / 5 s = 39,200 W

Now, to find the efficiency (η), we'll use the formula:

η = (P_out / P_in) * 100

Plugging in the values: η = (39,200 W / 500 W) * 100 ≈ 7840 %

Therefore, the efficiency of the motor in converting electrical power to mechanical power is approximately 7840%.