A motor that uses electrical energy at a rate of 1.5 kW is connected to a hydraulic lift that is used to lift a 1300 kg car to a height of 1.8 m in 24 s at a constant speed. Find the efficiency of the system consisting of the motor and the lift.

Question

shane_b · Answer

$$Work(J) = force \space *  \space distance = (1300kg)(9.8m/s^2)(1.8m)=22932J$$$$Power(W) = \frac{Joules}{second} = \frac{22932J}{24s}=955.5W=0.9555kW$$$$Eff(\%)=\frac{P_{out}}{P_{in}}*100=\frac{0.9555kW}{1.5kW}*100=63.7 \%$$