The 5 kg load is hoisted by pulley system and motor M. The load starts from rest and has speed of 2 m/s at the height of 1 m. If the motor has efficiency of e = 0.8, determine the power that must be applied to the motor at this instant.

1) 14.76 W.
2) 24.78 W.
3) 26.32 W.
4) 12.68 W.

The force that acts on the load does positive work, whereas the weight of the block does negative work.
Applying the work-energy equation,

The power output at this instant is given by-
P = F.v = 59.05(2) = 11.81 W.
The power input to the motor is,
power input = (power output)/e;
= 14.76W.