An 120 volt electric motor that draws 5 amps of current operates a winch that lifts a 100 kg. a distance of 300 cm. in 5 seconds.

Find work done by the winch, input power, output power, and efficiency of this system

Work done = 3000J

Input Power = 600W

Explanation:

Potential, V = 120 volt

Current, I = 5A

mass, m = 100 kg

distance, x = 300 cm = 3m

time, t = 5 sec

Work = ?

Work = Force X distance

Force needed to lift 100 kg object is 100 X 9.8 m/s = 1000N

So,

Work = 1000 x 3m = 3000J

Therefore, work done to lift 100 kg object is 3000J

Input power = ?

Input power = V x I

= 120 volt X 5A

= 600 W

Output power = ?

If the energy is conserved then,

Output power = input power = 600W

Efficiency = ?

Efficiency = output power / input power X 100

Efficiency = 600 X 100 / 600 = 100%

Therefore, no loss of energy