A spring is attached at one end to a fixed point and hangs vertically with a cube attached to

Q#22 (Past Exam Paper – November 2017 Paper 22 Q3)

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.

Fig. 3.1                                                            Fig. 3.2

The cube has weight 4.0 N and sides of length 5.1 cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0 cm below the water surface, as shown in Fig. 3.2.

The density of the water is 1000 kg m^-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 1.3 N. [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 30 N m^-1.

Determine the initial height above the water surface of the base of the cube before it was

released. [3]

(e) The cube in the water is released from the spring.

(i) Determine the initial acceleration of the cube. [2]

(ii) Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water. [2]

[Total: 12]

Solution:

(a)

{Pressure in liquid:      P = hρg}

{The top surface is at a depth of (7.0 – 5.1 =) 1.9 cm from the surface.

Calculate the pressure at the top surface and that at the bottom surface and find the difference.}

P = 1000 × 9.81 × 7.0 × 10^-2   or 1000 × 9.81 × 1.9 × 10^-2   

                  

ΔP = 1000 × 9.81 × (7.0 × 10^-2 – 1.9 × 10^-2) or 686 – 186

{Note that 7.0 cm – 1.9 cm is simply equal to the length of the cube.}

ΔP = 500 Pa                                                                                      

(b)

{Pressure = Force / Area       So, F = PA}

F = pA                         or (Δ)F = Δp × A

                                

upthrust = 500 × (5.1×10^-2)² = 1.3 N

or

upthrust = (686 – 186) × (5.1×10^-2)² = 1.3 N

or

upthrust = 1000 × 9.81 × 5.1×10^-2 × (5.1×10^-2)² = 1.3 N

(c)

{For equilibrium, upward forces = downward forces}

{Force exerted by spring + Upthrust = Weight

Force exerted by spring = Weight – Upthrust}

Force exerted by spring= 4.0 – 1.3 = 2.7 N

(d)

{Hooke’s law: F = ke}

extension e = 2.7 / 30

extension = 0.09 (m)   or 9 (cm)

{The extension is 9 cm. But the depth of the base beneath the surface is 7 cm. So,}

height above surface = 9 – 7 = 2 cm

(e)

(i)

{The spring no longer exerts any force on the cube.

Resultant force on cube = Weight – Upthrust = 4.0 – 1.3 = 2.7 N}

{Find the mass of the cube from W = mg}

mass = 4.0 / 9.81

{Resultant force = ma = 2.7 N}

Acceleration = 2.7 / (4.0 / 9.81) = 6.6 m s^-2

(ii)

{Since there is an acceleration, the speed increases. This causes the viscous force to increase. Thus the resultant force on the cube (and hence, the acceleration) decreases until it becomes zero [viscous force + upthrust = weight].

This is the same principle as for terminal velocity}

The viscous force increases (and then becomes constant)

(weight and upthrust constant so,) this causes the acceleration to decrease (to zero)

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