The total mass of a wheelbarrow and its load is 90kg. A worker pushes the wheelbarrow and load up a plank of length 2.0m on to a platform, as shown in Fig. 2.1. The platform is at a height of 0.60 m above the ground. [Figure 2.1] The worker exerts a force of 290 N on the wheelbarrow in the direction in which the wheelbarrow moves.
- (a)The gravitational field strength g is equal to 10N/kg.
- (a)(i)Calculate the gravitational potential energy gained by the wheelbarrow and its load. energy =[2]
- (a)(ii)The worker pushes the wheelbarrow 2.0m along the plank. Calculate the work done on the wheelbarrow by the worker. work done =[2]
- (a)(iii)Suggest one reason why the answer to (a)(ii) is not equal to the answer to (a)(i).[1]
- (b)The worker finds that walking up the plank pushing the wheelbarrow is even more inefficient than the answers in (a) suggest.
- (b)(i)State what is meant by efficiency.[1]
- (b)(ii)Suggest one reason why this method of lifting the load onto the platform is so inefficient.[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 (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] (a)(i) (g.p.e. =) mgh or 90 × 10 × 0.60
- [1] (a)(i) 540 J
- [1] (a)(ii) (W.D. =) Fx or 290 × 2.0
- [1] (a)(ii) 580 J
- [1] (a)(ii) friction (between wheel and axle) or friction (between tyre / wheel and plank) or (work done against) friction
- [1] (b)(i) useful energy output ( × 100%) or useful power output ( × 100%) (total) energy input (total) power input
- [1] (b)(ii) wheelbarrow / worker has to be lifted up ignore references to friction / air resistance
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