A wooden cubical block with a density of 710 kg/m^3 and a volume of 0.012 m^3 is

Question

A wooden cubical block with a density of 710 kg/m^3 and a volume of 0.012 m^3 is attached to the top of a vertical spring whose force constant is k = 540 N/m. (a) If the spring and the wooden block are completely immersed in water, will the spring be stretched or compressed? Find the amount by which the spring is stretched or compressed. (b) Assuming that, the spring is of negligible volume and it is cut: when the block settles, what Is the height of the block that is above the water? (rho_water = 1000 kg/m^3)

Explanation / Answer

(a) Forces on the block will be gravity which would act in downward direction and the buoyancy force which will act upward.
For time being we are not considering spring force.
Gravity = mg = (pV)*g
where p is density of block , V is volume of the block .
= 710*0.012*9.81 = 83.5812 N
Now buoynacy force = pwater*gV = 1000*9.81*0.012 = 117.72 N
hence we find that the buoyancy force is more than the weight hence the spring force must exerting force in downward direction to maintain the block with in the water.
That means spring is stretched.
Spring force = 117.72 - 83.5812 = 34.14 N
Spring force = Kx = 34.14
x = 34.14/540 = 0.063 m
hence the spring must be stretched by the 0.063 m
(b) When spring is cut then there would be no spring force then the weight should be balance by the buoynacy force.
Hence buoyancy force = pwgV = 83.5812
where V is the volume of the cube which is in water
V= 8.52*10-3 m3
Now we know that the block was cubical, therefore
volume = side3
0.012 = a3
a = 0.229
Therefore volume in the water = Area*height = a2*h = 8.52*10-3
h= 0.163 m
This is the height which will be in the water.
therefore height above water = a - h = 0.229 - 0.163 = 0.066 m
this height of the block must be above the water.

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