A-LevelPhysicsWork, energy and power9702/22 · Feb/Mar 2020 · Q312 Marks
State what is meant by work done.
- (a)State what is meant by work done.[1]
- (b)(i)A skier is pulled along horizontal ground by a wire attached to a kite, as shown in Fig. 3.1. [Figure 3.1] The skier moves in a straight line along the ground with a constant speed of 4.4ms⁻¹. The wire is at an angle of 30° to the horizontal. The tension in the wire is 140N. Calculate the work done by the tension to move the skier for a time of 30 s.[3]
- (b)(ii)The weight of the skier is 860N. The vertical component of the tension in the wire and the weight of the skier combine so that the skier exerts a downward pressure on the ground of 2400 Pa. Determine the total area of the skis in contact with the ground.[3]
- (b)(iii)The wire attached to the kite is uniform. The stress in the wire is 9.6 × 10⁶ Pa. Calculate the diameter of the wire.[2]
- (c)The variation with extension x of the tension F in the wire in (b) is shown in Fig. 3.2. [Figure 3.2] A gust of wind increases the tension in the wire from 140N to 210N. Calculate the change in the strain energy stored in the wire.[3]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- B1 force x displacement in the direction of the force
- C1 displacement = 4.4 × 30
- C1 work done = 140 cos 30° x 4.4 x 30
- A1 = 1.6 × 10⁴ J
- C1 p = F/A
- C1 F = 860 – 140 sin 30° (= 790)
- A1 A = 790 / 2400
- A1 = 0.33 m²
- C1 σ = F/A or F/ πr² or 4F / πd²
- C1 9.6 x 10⁶ = 4 × 140 / πd²
- A1 d = 4.3 x 10⁻³ m
- C1 E = ½Fx or ½kx² or area under graph
- C1 (Δ)Ε = ½ × (140 + 210) × 0.20 × 10⁻³
- C1 or
- C1 (Δ)Ε = (½ × 210 × 0.60 × 10⁻³) – (½ × 140 × 0.40 × 10⁻³)
- C1 or
- C1 (Δ)Ε = (140 × 0.20 × 10⁻³) + (½ x 0.20 × 10⁻³ × 70)
- C1 or
- C1 (Δ)Ε = [½×3.5 × 10⁵ × (0.60 × 10⁻³)² ] – [½ x 3.5 × 10⁵ × (0.40 × 10⁻³)² ]
- A1 ΔΕ = 0.035 J
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