A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end. The cube is initially held so that the spring has zero extension, as shown in Fig. 3.1. [Figure 3.1] The cube has weight 4.0N and sides of length 5.1cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0cm below the water surface, as shown in Fig. 3.2. [Figure 3.2] The density of the water is 1000kgm⁻³.
- (a)Calculate the difference in the pressure exerted by the water on the bottom face and on the top face of the cube.[2]
- (b)Use your answer in (a) to show that the upthrust on the cube is 1.3N.[2]
- (c)Calculate the force exerted on the spring by the cube when it is in equilibrium in the water.[1]
- (d)The spring obeys Hooke's law and has a spring constant of 30Nm⁻¹. Determine the initial height above the water surface of the base of the cube before it was released.[3]
- (e)(i)The cube in the water is released from the spring. Determine the initial acceleration of the cube.[2]
- (e)(ii)Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water.[2]
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Mark scheme
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- [C1] ∆p = 1000 × 9.81 × 7.0 × 10⁻² or 1000 × 9.81 × 1.9 × 10⁻² or ∆p = 1000 × 9.81 × (7.0 × 10⁻² – 1.9 × 10⁻²) or 686 – 186
- [A1] = 500 Pa
- [C1] F = pA or (∆)F = ∆p × A
- [A1] upthrust = 500 × (5.1 × 10⁻²)² = 1.3N or upthrust = (686 – 186) × (5.1 ×10⁻²)² = 1.3N or upthrust = 1000 × 9.81 × 5.1 ×10⁻² × (5.1 × 10⁻²)² = 1.3 N
- [A1] force = 4.0 – 1.3 = 2.7 N
- [C1] extension/x/e = 2.7/30 = 0.09 (m) or 9 (cm)
- [C1] height above surface = 9 – 7
- [A1] = 2 cm
- [C1] mass = 4.0/9.81
- [A1] acceleration = 2.7/(4.0/9.81) = 6.6 ms⁻²
- [M1] viscous force increases (and then becomes constant)
- [A1] (weight and upthrust constant so) acceleration decreases (to zero)
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