A double pulley system is used to lift a 300 lb box. The motor pulls the top rope, r₁ at a velocity of 6 in/s. The motor efficiency is 0.4. What is the require power rating in horsepower for the motor? Ignore the weight of the ropes and pulleys.

Position when v = 0 and a = 0

The motor draws in rope 1, r₁, at a speed of 6 in/s (0.5 ft/s). A single pulley will reduce the rope speed by half, giving the velocity of r₂ and r₃ as,
     v₂ = v₁/2 = 0.25 ft/s
     v₃ = v₂/2 = 0.125 ft/s

Thus the power output needed from the motor is,
      P_out = F·v₃
             = 300(0.125) = 37.5 ft-lb/s

The motor efficiency determines the power input of the motor,
     η = P_out/P_in
     0.4 = 37.5/P_in
     P_in = 93.75 ft-lb/s

Convert to horsepower
     93.75/550 = 0.1705 hp

P_in = 0.1705 hp