Fig. 2.1 illustrates an electric motor lifting a load. The load, which has a weight of $5.0\,\text{N}$, is lifted vertically through $3.5\,\text{m}$ at constant speed. The efficiency of the electric motor is $0.65$ (65%).
(a)[2]
Calculate the rise in gravitational potential energy of the load.
(b(i))[1]
State the formula connecting efficiency, energy input and useful energy output.
(b(ii))[2]
Calculate the energy input to the motor.
(c)[1]
Suggest one reason why the efficiency of the motor is less than $1.0$ (100%).
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “PE = mgh, or Fd, or 5 \times 3.5” …