O-LevelPhysicsEnergy & Work5054/22 · May/June 2016 · Q26 Marks

Fig. 2.1 shows an electric motor lifting a load. The load, of weight 5.0N, is raised through a vertical height of 3.5m at a constant speed. The efficiency of the electric motor is 0.65 (65%).

(a)Calculate the increase in gravitational potential energy of the load.[2]
(b)(i)State the formula that relates efficiency, energy input and useful energy output.[1]
(b)(ii)Calculate the energy input to the motor.[2]
(c)Suggest one reason why the efficiency of the motor is less than 1.0 (100%).[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

[1] (PE =) mgh or Fd or 5 × 3.5
[1] 17.5 J or 17 J or 18 J
[1] (efficiency = useful) energy output/energy input in any form but all three quantities must be mentioned if efficiency is not the subject of the equation
[1] 17.5/0.65 or 17.5/65 or 0.65/65 = (a)/energy input
[1] 26.9 J or 27 J
[1] due to friction (in bearings of motor) or due to (electrical) resistance (in motor) or air resistance acts or thermal energy/heat produced/lost (in resistance of motor/due to friction)

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.