The variation with extension x of the force F applied to a spring is shown in Fig. 4.1. [Figure 4.1] The spring has an unstretched length of 0.080 m and is suspended vertically from a fixed point, as shown in Fig. 4.2. [Figure 4.2] A block is attached to the lower end of the spring. The block hangs in equilibrium at position X when the length of the spring is 0.095 m, as shown in Fig. 4.3. [Figure 4.3] The block is then pulled vertically downwards and held at position Y so that the length of the spring is 0.120m, as shown in Fig. 4.4. The block is then released and moves vertically upwards from position Y back towards position X. [Figure 4.4]
- (a)Use Fig. 4.1 to determine the spring constant of the spring. spring constant = ............................................... Nm⁻¹ [2][2]
- (b)Use Fig. 4.1 to show that the decrease in elastic potential energy of the spring is 0.055 J when the block moves from position Y to position X. [2][2]
- (c)The block has a mass of 0.122kg. Calculate the increase in gravitational potential energy of the block for its movement from position Y to position X. increase in gravitational potential energy = ................................................. J [2][2]
- (d)Use the decrease in elastic potential energy stated in (b) and your answer in (c) to determine, for the block, as it moves through position X:
- (d)(i)its kinetic energy kinetic energy = ................................................. J [1][1]
- (d)(ii)its speed. speed = .................................................... ms⁻¹ [2][2]
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 2019). The mark or mark type is shown in brackets.
- [C1] k = F/x or k = gradient; e.g. k = 4.0/0.050
- [A1] k = 80 N m⁻¹
- [C1] E = ½Fx or E = ½kx² or E = area under graph
- [A1] (Δ)Ε = (½ × 3.2 × 0.040) – (½ x 1.2 x 0.015) = 0.055 J; or (Δ)Ε = (½ × 80 × 0.040²) – (½ × 80 × 0.015²) = 0.055 J; or (Δ)Ε = ½ × (1.2 + 3.2) × 0.025 = 0.055 J
- [C1] (Δ)Ε = mg(Δ)h; = 0.122 x 9.81 × (0.120 – 0.095)
- [A1] = 0.030 J
- [(C1)] (Δ)Ε = W × (Δ)h; = 1.2 × 0.025
- [(A1)] = 0.030 J
- [A1] E = 0.055 -0.030; = 0.025 J
- [C1] E = ½mv²; V = [(2 × 0.025) / 0.122]⁰.⁵
- [A1] = 0.64 m s⁻¹
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