Question 22
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]
Reference: Past Exam Paper – November 2017 Paper 22 Q3
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)
i dont get d) how is it 2cm?
ReplyDeleteAs calculated using Hooke’s law, the extension is 9 cm. So previously, the base was 9 cm above the position shown in the diagram. But the depth of the base beneath the surface is 7 cm – that when raised by 7 cm, the base would then be at the surface of the water. So, it still needs to be raised by the remaining 2 cm.
DeleteWhy do we not use F as the weight of the cube? Isn’t 2.7 the force exerted when in equilibrium with water? Or does it remain constant throughout? :/ even if it isn’t in equilibrium
ReplyDeleteThe weight of the cube is given to be 4.0 N. This value of weight is constant on earth and at any position in the water.
DeleteF in F = ke is the force in the spring. This has a value of 2.7 N when in equilibrium. However if the spring is not fully immersed in the water, this value may change.