A-LevelPhysicsWork, energy and power9702/23 · Oct/Nov 2019 · Q29 Marks

A lift (elevator) of weight 13.0kN is connected by a cable to a motor, as shown in Fig. 2.1. [Figure 2.1] The lift is pulled up a vertical shaft by the cable. A constant frictional force of 2.0kN acts on the lift when it is moving. The variation with time t of the speed v of the lift is shown in Fig. 2.2. [Figure 2.2]

(a)State what is meant by work done.[1]
(b)(i) 1.the acceleration of the lift between time t = 0 and t = 3.0 s[2]
(b)(i) 2.the work done by the motor to raise the lift between time t = 3.0s and t = 6.0s.[2]
(b)(ii)The motor has an efficiency of 67%. The tension in the cable is 1.6 × 104 N at time t = 2.5 s. Determine the input power to the motor at this time.[3]
(b)(iii)State and explain whether the increase in gravitational potential energy of the lift from time t = 0 to t = 7.0s is less than, the same as, or greater than the work done by the motor. A calculation is not required.[1]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] (work done =) force × distance moved in direction of force
[1] acceleration = gradient or a = (v – u)/t or a = ∆v/t
[1] e.g. a = 2.4/3.0 = 0.80 m s-2
[1] tension in cable = (13.0 + 2.0) × 103
[1] work done = 15 × 103 × (3.0 × 2.4) = 1.1 x 105 J
[1] power = Fv
[1] v = 2.0 (m s⁻¹)
[1] input power = (1.6 × 104 × 2.0)/0.67 = 4.8 × 104 W
[1] work is done against friction so (increase in) GPE is less (than work done by motor) or energy is lost or transferred or converted to heat/thermal energy due to friction or resistance force or work is done lifting the cable so GPE is less

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