Calculate the useful power output for an electric motor that uses 200 W and has an
efficiency of 40%

Correct option is B)Power with which crane lifts the load =0.8×0.9 × input powerP=0.8×0.9×200 kWNow, P= F . v Here, F=mg (since there is no acceleration)10 3 ×0.8×0.9×200=3.6×10 3 ×10×v⇒v=4 m/s